【MOOC課程學習記錄】程序設計與算法（一）C語言程序設計

時間 2020-06-03

原文原文鏈接

課程結課了，把作的習題都記錄一下，告訴本身多少學了點東西，也能給本身一點鼓勵。ios

ps：題目都在cxsjsxmooc.openjudge.cn上能看到，參考答案在差很少結課的時候也會在mooc上放出來。c++

程序設計與算法（一）C語言程序設計第一週習題(2019夏季)算法

001:輸出第二個整數編程

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	int a, b, c;
	cin >> a >> b >> c;
	cout << b;
	return 0;
}

002:字符菱形ide

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	char c;
	scanf("%c", &c);
	printf("  %c\n", c);
	printf(" %c%c%c\n", c, c, c);
	printf("%c%c%c%c%c\n", c, c, c, c, c);
	printf(" %c%c%c\n", c, c, c);
	printf("  %c\n", c);
	return 0;
}

003:打印ASCII碼函數

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	char c;
	scanf("%c", &c);
	printf("%d\n", c);
	return 0;
}

004:打印字符測試

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	int c;
	scanf("%d", &c);
	printf("%c\n", c);
	return 0;
}

005:整型數據類型存儲空間大小編碼

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	int a;
	short b;
	printf("%lu ", sizeof(a));
	printf("%lu", sizeof(b));
	return 0;
}

006:浮點型數據類型存儲空間大小加密

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	float a;
	double b;
	printf("%lu ", sizeof(a));
	printf("%lu", sizeof(b));
	return 0;
}

程序設計與算法（一）C語言程序設計第二週習題(2019夏季)spa

001:對齊輸出

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	int a, b, c;
	scanf("%d %d %d", &a, &b, &c);
	printf("%8d %8d %8d", a, b, c);
	return 0;
}

002:輸出保留12位小數的浮點數

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	double a;
	scanf("%lf", &a);
	printf("%.12lf", a);
	return 0;
}

003:空格分隔輸出

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	char a;
	int b;
	float c;
	double d;
	scanf("%c", &a);
	scanf("%d", &b);
	scanf("%f", &c);
	scanf("%lf", &d);
	printf("%c ", a);
	printf("%d ", b);
	printf("%.6f ", c);
	printf("%.6lf", d);
	return 0;
}

004:計算球的體積

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	double d;
	scanf("%lf", &d);
	printf("%.2lf", 4.0/3*3.14*d*d*d);
	return 0;
}

005:大象喝水

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	int h, r;
	float c;
	scanf("%d %d", &h, &r);
	c = 3.14159*r*r*h;
	printf("%d", int(20000/c+1));
	return 0;
}

程序設計與算法（一）C語言程序設計第三週習題(2019夏季)

001:奇偶數判斷

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	int a;
	cin >> a;
	if (a % 2){
		cout << "odd";
	}else{
		cout << "even";
	}
	return 0;
}

002:求一元二次方程的根

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

#define EPS 1e-6

int main(){
	double a, b, c;
	scanf("%lf %lf %lf", &a, &b, &c);
	double x1, x2, tmp;
	if (b*b-4*a*c > -EPS && b*b-4*a*c < EPS){
		x1 = x2 = -b/(2*a);
		printf("x1=x2=%.5lf", x1);
	}else if (b*b-4*a*c >= EPS){
		x1 = ((-b)+sqrt(b*b-4*a*c))/(2*a);
		x2 = ((-b)-sqrt(b*b-4*a*c))/(2*a);
		if (x1 < x2){
			tmp = x1;
			x1 = x2;
			x2 = tmp;
		}
		printf("x1=%.5lf;x2=%.5lf", x1, x2);
	}else{
		// 此處x1用來儲存實部係數，x2用來儲存虛部係數
		x1 = -b/(2*a);
		x2 = sqrt(4*a*c-b*b)/(2*a);
		if (x1 < EPS && x1 > -EPS){
			x1 = 0;
		}
		printf("x1=%.5lf+%.5lfi;x2=%.5lf-%.5lfi", x1, x2, x1, x2);
	}
	return 0;
}

003:點和正方形的關係

#include <iostream>
#include <cstdio>
using namespace std;


int main(){
	int x, y;
	scanf("%d %d", &x, &y);
	if (x <= 1 && x >= -1 && y <= 1 && y >= -1){
		printf("yes");
	}else{
		printf("no");
	}
	return 0;
}

004:蘋果和蟲子2

#include <iostream>
#include <cstdio>
using namespace std;


int main(){
	int n, x, y;
	cin >> n >> x >> y;
	int a;
	if (y%x){
		a = n-y/x-1;
	}else{
		a = n-y/x;
	}
	if (a < 0){
		a = 0;
	}
	cout << a;
	return 0;
}

005:簡單計算器

#include <iostream>
#include <cstdio>
using namespace std;


int main(){
	int opd1, opd2;
	char opt;
	scanf("%d %d %c", &opd1, &opd2, &opt);
	int res;
	switch (opt){
	case '+':
		res = opd1+opd2;
		printf("%d", res);
		break;
	case '-':
		res = opd1-opd2;
		printf("%d", res);
		break;
	case '*':
		res = opd1*opd2;
		printf("%d", res);
		break;
	case '/':
		if (!opd2){
			printf("Divided by zero!");
		}else{
			res = opd1/opd2;
			printf("%d", res);
		}
		break;
	default:
		printf("Invalid operator!");
		break;
	}
	return 0;
}

006:求整數的和與均值

#include <iostream>
#include <cstdio>
using namespace std;


int main(){
	int n, sum;
	double ave;
	sum = 0;
	cin >> n;
	int i, tmp;
	for (i = 0; i < n; i++){
		cin >> tmp;
		sum += tmp;
	}
	ave = 1.0*sum/n;
	printf("%d %.5f", sum, ave);
	return 0;
}

007:整數序列的元素最大跨度值

#include <iostream>
#include <cstdio>
using namespace std;


int main(){
	int n;
	cin >> n;
	int max, min;
	max = 0;
	min = 1000;
	int i, tmp;
	for (i = 0; i < n; i++){
		cin >> tmp;
		if (tmp > max){
			max = tmp;
		}
		if (tmp < min){
			min = tmp;
		}
	}
	cout << max - min;
	return 0;
}

008:奧運獎牌計數

#include <iostream>
#include <cstdio>
using namespace std;


int main(){
	int n;
	scanf("%d", &n);
	int goldMedal, silverMedal, bronzeMedal;
	goldMedal = silverMedal = bronzeMedal = 0;
	int i, a, b, c;
	for (i = 0; i < n; i++){
		scanf("%d %d %d", &a, &b, &c);
		goldMedal += a;
		silverMedal += b;
		bronzeMedal += c;
	}
	int totalMedal;
	totalMedal = goldMedal+silverMedal+bronzeMedal;
	printf("%d %d %d %d", goldMedal, silverMedal, bronzeMedal, totalMedal);
	return 0;
}

009:乘方計算

#include <iostream>
#include <cstdio>
using namespace std;


int main(){
	int a, n, res;
	res = 1;
	scanf("%d %d", &a, &n);
	while (n > 0){
		res *= a;
		n--;
	}
	printf("%d", res);
	return 0;
}

010:雞尾酒療法

#include <iostream>
#include <cstdio>
using namespace std;


int main(){
	int n;
	cin >> n;
	double a[n];
	int i, total, effective;
	for (i = 0; i < n; i++){
		cin >> total >> effective;
		a[i] = 1.0*effective/total;
	}
	for (i = 1; i < n; i++){
		if (a[i]-a[0] > 0.05){
			cout << "better\n";
		}else if (a[i]-a[0] < -0.05){
			cout << "worse\n";
		}else{
			cout << "same\n";
		}
	}
	return 0;
}

程序設計與算法（一）C語言程序設計第四周習題(2019夏季)

001:角谷猜測

#include <iostream>
#include <cstdio>
using namespace std;


int main(){
	long long n;
	scanf("%lld", &n);
	while (n > 1){
		if (n%2){
			printf("%lld*3+1=%lld\n", n, 3*n+1);
			n = 3*n+1;
		}else{
			printf("%lld/2=%lld\n", n, n/2);
			n /= 2;
		}
	}
	printf("End");
	return 0;
}

002:正常血壓

#include <iostream>
#include <cstdio>
using namespace std;


int main(){
	int n;
	cin >> n;
	int max, tmp, systolicPressure, diastolicPressure;
	max = tmp = 0;
	int i;
	for (i = 0; i < n; i++){
		cin >> systolicPressure >> diastolicPressure;
		if (systolicPressure <= 140 && systolicPressure >= 90
		&& diastolicPressure <= 90 && diastolicPressure >= 60){
			tmp++;
			if (tmp > max){
				max = tmp;
			}
		}else{
			tmp = 0;
		}
	}
	cout << max;
	return 0;
}

003:數字反轉

#include <iostream>
#include <cstdio>
using namespace std;

int main(int argc, char const *argv[])
{
    int n;
    scanf("%d", &n);
    int sign = 1;
    if (n < 0){
        sign = -1;
        n = -n;
    }
    int rev = 0;
    while (n > 0){
        rev = rev*10 + n%10;
        n /= 10;
    }
    rev = sign*rev;
    printf("%d", rev);
    return 0;
}

004:求特殊天然數

#include <iostream>
#include <cstdio>
using namespace std;

int main(int argc, char const *argv[])
{
    int n10, n7, n9;
    int a[3], b[3], tmpA, tmpB, i, j;
    for (n10 = 81; n10 < 343; n10++){
        tmpA = tmpB = n10;
        i = j = 2;
        while (tmpA > 0){
            a[i] = tmpA%7;
            tmpA /= 7;
            i--;
        }
        while (tmpB > 0){
            b[j] = tmpB%9;
            tmpB /= 9;
            j--;
        }
        if (a[0] == b[2] && a[1] == b[1] && a[2] == b[0]){
            break;
        }
    }
    cout << n10 << "\n" << a[2]+a[1]*10+a[0]*100 << "\n" << b[2]+b[1]*10+b[0]*100;
    return 0;
}

005:僱傭兵

#include <iostream>
#include <cstdio>
using namespace std;

#define MAXSIZE 10

int main(int argc, char const *argv[])
{
    int M, N, X;
    cin >> M >> N >> X;
    while (N*X >= M){
        X -= M/N+1;
        N += M/N;
    }
    printf("%d", N);
    return 0;
}

006:數字統計

#include <iostream>
#include <cstdio>
using namespace std;

int main(int argc, char const *argv[])
{
    int a, b;
    cin >> a >> b;
    int i, tmp, counter;
    counter = 0;
    for (i = a; i <= b; i++){
        int t = i;
        while (t>0){
            tmp = t%10;
            t /= 10;
            if (tmp == 2){
                counter++;
            }
        }
    }
    cout << counter;
    return 0;
}

程序設計與算法（一）C語言程序設計第五週習題(2019夏季)

001:與指定數字相同的數的個數

#include <iostream>
#include <cstdio>
using namespace std;
    
int main(int argc, char const *argv[])
{
    int n;
    scanf("%d", &n);
    int a[n], i;
    for (i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }
    int specifiedNumber;
    scanf("%d", &specifiedNumber);
    int counter = 0;
    for (i = 0; i < n; i++){
        if (a[i] == specifiedNumber){
            counter++;
        }
    }
    printf("%d", counter);
    return 0;
}

002:陶陶摘蘋果

#include <iostream>
#include <cstdio>
using namespace std;

#define MAXSIZE 10

int main(int argc, char const *argv[])
{
    int height[MAXSIZE];
    int i;
    int counter = 0;
    for (i = 0; i < MAXSIZE; i++){
        cin >> height[i];
    }
    int taotao;
    int bench = 30;
    cin >> taotao;
    for (i = 0; i < 10; i++){
        if (height[i] <= taotao + bench){
            counter++;
        }
    }
    cout << counter;
    return 0;
}

003:年齡與疾病

#include <iostream>
#include <cstdio>
using namespace std;

#define MAXSIZE 10

int main(int argc, char const *argv[])
{
    int n;
    scanf("%d", &n);
    int a[n], i, counter18, counter35, counter60, counterOther;
    counter18 = counter35 = counter60 = counterOther = 0;
    for (i = 0; i < n; i++){
        scanf("%d", &a[i]);
        if (a[i] <=18 ){
            counter18++;
        }else if (a[i] <= 35){
            counter35++;
        }else if (a[i] <= 60){
            counter60++;
        }else{
            counterOther++;
        }
    }
    printf("%.2f%%\n", 100.0*counter18/n);
    printf("%.2f%%\n", 100.0*counter35/n);
    printf("%.2f%%\n", 100.0*counter60/n);
    printf("%.2f%%\n", 100.0*counterOther/n);
    return 0;
}

004:校門外的樹

#include <iostream>
#include <cstdio>
using namespace std;

int main(int argc, char const *argv[])
{
    int L, M;
    scanf("%d %d", &L, &M);
    int a[L+1], tmpA, tmpB, i;
    for (i = 0; i < L+1; a[i++] = 1)
        ;
    for (i = 0; i < M; i++){
        scanf("%d %d", &tmpA, &tmpB);
        int j;
        for (j = tmpA; j <= tmpB; a[j++] = 0)
            ;
    }
    int counter = 0;
    for (i = 0; i < L+1; i++){
        if (a[i] == 1){
            counter++;
        }
    }
    printf("%d", counter);
    return 0;
}

005:計算鞍點

#include <iostream>
#include <cstdio>
using namespace std;

#define MAXLINE 5
#define MAXROW 5

int main(int argc, char const *argv[])
{
    int a[MAXLINE][MAXROW], i, j, lmax, max[MAXLINE];
    for (i = 0; i < MAXLINE; i++){
        cin >> a[i][0];
        lmax = a[i][0];
        max[i] = 0;
        for (j = 1; j < MAXROW; j++){
            cin >> a[i][j];
            if (a[i][j] > lmax){
                lmax = a[i][j];
                max[i] = j;
            }
        }
    }
    int found = 1;
    for (i = 0; i < MAXLINE; i++){
        for (j = 0; j < MAXLINE; j++){
            if (a[j][max[i]] < a[i][max[i]]){
                found = 0;
            }
        }
        if (found){
            break;
        }else if (i == MAXLINE-1){
            break;
        }
        else{
            found = 1;
        }
    }
    if (found){
        cout << i+1 << " " << max[i]+1 << " " << a[i][max[i]];
    }else{
        cout << "not found";
    }
    
    return 0;
}

006:圖像模糊處理

#include <iostream>
#include <cstdio>
using namespace std;

int main(int argc, char const *argv[])
{
    int m, n;
    scanf("%d %d", &m, &n);
    int a[m][n], b[m][n], i, j;
    for (i = 0; i < m; i++){
        for (j = 0; j < n; j++){
            scanf("%d", &a[i][j]);
            b[i][j] = a[i][j];
        }
    }
    for (i = 1; i < m-1; i++){
        for (j = 1; j < n-1; j++){
            b[i][j] = (2*(a[i][j-1]+a[i][j+1]+a[i][j]+a[i-1][j]+a[i+1][j])+5)/10;
        }
    }
    for (i = 0; i < m; i++){
        for (j = 0; j < n; j++){
            printf("%d ", b[i][j]);
        }
        printf("\n");
    }
    return 0;
}

007:矩陣轉置

#include <iostream>
#include <cstdio>
using namespace std;

int main(int argc, char const *argv[])
{
    int m, n;
    cin >> m >> n;
    int a[m][n], i, j;
    for (i = 0; i < m; i++){
        for (j = 0; j < n; j++){
            cin >> a[i][j];
        }
    }
    for (j = 0; j < n; j++){
        for (i = 0; i < m; i++){
            cout << a[i][j] << " ";
        }
        cout << endl;
    }
    return 0;
}

程序設計與算法（一）C語言程序設計第六週習題(2019夏季)

001:Pell數列

#include <iostream>
#include <cstdio>
using namespace std;

int main(int argc, char const *argv[])
{
    int n;
    scanf("%d", &n);
    int a[1000000];
    a[0] = 1;
    a[1] = 2;
    int i;
    for (i = 2; i < 1000000; i++){
        a[i] = 2*a[i-1]+a[i-2];
        if (a[i] > 32767){
            a[i] %= 32767;
        }
    }
    int b[n];
    for (i = 0; i < n; i++){
        scanf("%d", &b[i]);
    }
    for (i = 0; i < n; i++){
        printf("%d\n", a[b[i]-1]%32767);
    }
    return 0;
}

002:求最大公約數問題

#include <iostream>
#include <cstdio>
using namespace std;

int main(int argc, char const *argv[])
{
    int a, b, tmp;
    cin >> a >> b;
    do{
        tmp = a%b;
        a = b;
        b = tmp;
    }while (tmp);
    cout << a;
    return 0;
}

003:編程填空：第i位替換

return 1<<i&m|~(1<<i)&n;

004:編程填空：第i位取反

return 1<<i^n;

005:編程填空：左邊i位取反

return ~0<<32-i^n;

程序設計與算法（一）C語言程序設計第七週習題(2019夏季)

001:統計數字字符個數

#include <iostream>
#include <cstdio>
using namespace std;

int main(int argc, char const *argv[])
{
    int counter = 0, c;
    while ((c = getchar()) != '\n'){
        if (c >= '0' && c <= '9'){
            counter++;
        }
    }
    cout << counter;
    return 0;
}

002:找第一個只出現一次的字符

#include <iostream>
#include <cstdio>
using namespace std;

char a[100000];
int b[26];

int main(int argc, char const *argv[])
{   
    int i;
    for (i = 0; i < 26; i++){
        b[i] = 0;
    }
    cin >> a;
    // int c;
    // while ((c = getchar()) != '\n'){
    //     b[c-'a']++;
    // }
    for (i = 0; a[i]; i++){
        b[a[i]-'a']++;
    }
    for (i = 0; a[i]; i++){
        if (b[a[i]-'a'] == 1){
            cout << a[i];
            break;
        }
    }
    if (a[i] == '\0'){
        cout << "no";
    }
    return 0;
}

003:石頭剪子布

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int stoi(char *s);

int main(int argc, char const *argv[])
{   
    char player1S[10], player2S[10];
    int player1I, player2I, N;
    scanf("%d", &N);
    int i, a[N];
    for (i = 0; i < N; i++){
        scanf("%s %s", player1S, player2S);
        player1I = stoi(player1S);
        player2I = stoi(player2S);
        int t = player1I - player2I;
        if (t == 0){
            a[i] = 0;
        }else if (t == -1 || t == 2){
            a[i] = 1;
        }else{
            a[i] = -1;
        }
    }
    for (i = 0; i < N; i++){
        if (a[i] == -1){
            printf("%s\n", "Player2");
        }else if (a[i] == 1){
            printf("%s\n", "Player1");
        }else{
            printf("%s\n", "Tie");
        }
    }
    return 0;
}

int stoi(char *s){
    int p;
    if (!strcmp(s, "Rock"))
        p = -1;
    else if (!strcmp(s, "Scissors"))
        p = 0;
    else
        p = 1;
    return p;
}

004:最長最短單詞

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

#define MAXSIZE 20000
#define MAXWORDSIZE 100
#define IN 1
#define OUT 0

char s[MAXSIZE];
char longest[MAXWORDSIZE];
char shortest[MAXWORDSIZE];
char tmp[MAXWORDSIZE];

int main(int argc, char const *argv[])
{   
    gets(s);
    int state = OUT, max = -1, min = MAXWORDSIZE+1, len;
    int i = 0, j = 0;
    while (1){
        if (s[i] == ' ' || s[i] == ',' || !s[i]){
            if (state == IN){
                tmp[j] = '\0';
                j = 0;
                len = strlen(tmp);
                if (len > max){
                    max = len;
                    strcpy(longest, tmp);
                }
                if (len < min){
                    min = len;
                    strcpy(shortest, tmp);
                }
                if (!s[i]){
                    break;
                }
            }else{
                i++;
                continue;
            }
            state = OUT;
            i++;
        }
        tmp[j++] = s[i++];
        state = IN;
    }
    printf("%s\n", longest);
    printf("%s\n", shortest);
    return 0;
}

005:密碼翻譯

#include <iostream>
#include <cstdio>
using namespace std;

#define MAXSIZE 80

char s[MAXSIZE];

int main(int argc, char const *argv[])
{   
    cin.getline(s, MAXSIZE);
    int i;
    for (i = 0; s[i]; i++){
        if (s[i] == 'z'){
            s[i] = 'a';
        }else if (s[i] == 'Z'){
            s[i] = 'A';
        }else if (s[i] >= 'a' && s[i] <= 'y' || s[i] >= 'A' && s[i] <= 'Y'){
            s[i] += 1;
        }
    }
    cout << s;
    return 0;
}

程序設計與算法（一）C語言程序設計第8、九周習題(2019夏季)

001:指針練習：輸出Hello

p = s; *p; p++

002:指針練習:輸出Tesla

; *p1 != *p2; p1++

003:指針練習：ForEach

void (*f)(void *)

004:指針練習：Memcpy之一

int i;
    for (i = 0; i < n; i++){
        dest[i] = src[i];
    }

005:指針練習：double

a[1], 6

006:指針練習：Memcpy之二

char *sourceCode = (char*)src;
    char *destination = (char*)dest;
    char b[size], i;
    for (i = 0; i < size; i++){
        b[i] = *sourceCode++;
    }
    char *tmp = b;
    for (i = 0; i < size; i++){
        *destination++ = *tmp++;
    }

007:指針練習：MyMax

void *MyMax(void *anArray, int sizeOfAnItem, int numberOfItem, 
    int (*comparisonFunction)(void *n1, void*n2)){
        char *tmp = (char*)anArray;
        int i;
        void *max = anArray;
        for (i = 0; i < numberOfItem; i++){
            if (comparisonFunction(tmp, max) > 0){
                max = tmp;
            }
            tmp = tmp+sizeOfAnItem;
        }
    return max;
}

008:指針練習：指向指針的指針

int **p = a;

009:指針練習：SwapMemory

char *a1 = (char*)m1;
    char *a2 = (char*)m2;
    char c;
    int i;
    for (i = 0; i < size; i++){
        c = *a1;
        *a1++ = *a2;
        *a2++ = c;
    }

程序設計與算法（一）C語言程序設計第十週習題(2019夏季)

001:成績排序

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

struct student{
	char name[20];
	int score;
};

int main(){
	int n;
	scanf("%d", &n);
	int i;
	struct student a[20];
	for (i = 0; i < n; i++){
		scanf("%s %d", &a[i].name, &a[i].score);
	}
	struct student tmp;
	int j, index;
	for (i = 0; i < n; i++){
		tmp = a[i];
		index = i;
		for (j = i; j < n; j++){
			if (a[j].score > tmp.score){
				tmp = a[j];
				index = j;
			}else if (a[j].score == tmp.score){
				if (strcmp(tmp.name, a[j].name) > 0){
					tmp = a[j];
					index = j;
				}
			}
		}
		a[index] = a[i];
		a[i] = tmp;
	}
	for (i = 0; i < n; i++){
		printf("%s %d\n", a[i].name, a[i].score);
	}
	return 0;
}

002:分數線劃定

#include <iostream>
#include <cstdio>

using namespace std;

struct candidate{
	int ID;
	int score;
};

int main(){
	int n, m;
	scanf("%d %d", &n, &m);
	int i;
	struct candidate a[n];
	for (i = 0; i < n; i++){
		scanf("%d %d", &a[i].ID, &a[i].score);
	}
	struct candidate tmp;
	int j, index;
	for (i = 0; i < n; i++){
		tmp = a[i];
		index = i;
		for (j = i; j < n; j++){
			if (a[j].score > tmp.score){
				tmp = a[j];
				index = j;
			}else if (a[j].score == tmp.score){
				if (a[j].ID < tmp.ID){
					tmp = a[j];
					index = j;
				} 
			}
		}
		a[index] = a[i];
		a[i] = tmp;
	}
	
	int t = m*3/2-1;
	tmp = a[t];
	int boundary = a[t].score;
	while (tmp.score == boundary){
		t++;
		tmp = a[t];
	}
	printf("%d %d\n", boundary, t);
	for (i = 0; i < t; i++){
		printf("%d %d\n", a[i].ID, a[i].score);
	}
	return 0;
}

003:病人排隊

#include <iostream>
#include <cstdio>

#define BOUNDARY 60

using namespace std;

struct patient{
	char ID[10];
	int age;
};

int main(){
	int n;
	scanf("%d", &n);
	int i;
	struct patient a[n];
	for (i = 0; i < n; i++){
		scanf("%s %d", &a[i].ID, &a[i].age);
	}
	struct patient tmp;
	int j, index;
	for (i = 0; i < n; i++){
		for (j = n-1; j > i; j--){
			if (a[j].age >= BOUNDARY){
				if (a[j-1].age < a[j].age){
					tmp = a[j];
					a[j] = a[j-1];
					a[j-1] = tmp;			
				}
			}
		}
	}
	
	for (i = 0; i < n; i++){
		printf("%s\n", a[i].ID);
	}
	return 0;
}

004:mysort

void mysort(void *t, int numberOfElement, unsigned int widthOfElement, 
	int(*fCompare)(const void *, const void *)){
		int i, j, k;
		char *p = (char*)t;
		char tmp;
		for (i = 0; i < numberOfElement; i++){
			for (j = numberOfElement-1; j > 0; j--){
				char *a, *b;
				a = p+(j)*widthOfElement;
				b = p+(j-1)*widthOfElement;
				if (fCompare(a, b) < 0){
					for (k = 0; k < widthOfElement; k++){
						tmp = *a;
						*a++ = *b;
						*b++ = tmp;
					}
				}
			}
		}
		return;
}

005:從字符串中取數

static char *start;
    if (str){
        start = str;
    }
    double f = 0;;
    for (; *start < '0' || *start > '9'; start++){
        if (*start == '\0'){
            return -1;
        }
    }
    for (; *start >= '0' && *start <= '9'; start++){
        f = 10*f+(*start-'0');
    }
    if (*start == '.'){
        int tmp = 1;
        start++;
        for (; *start >= '0' && *start <= '9'; start++){
            f = 10*f+(*start-'0');
            tmp *= 10;
        }
        f /= tmp;
    }
    return f;

程序設計與算法（一）C語言程序設計第十一週習題(2019夏季)

001:派

#include <iostream>
#include <cstdio>

#define PI 3.141592653589793
#define EPS 0.00001 

using namespace std;

int main ()
{	
	int N, F;
	scanf("%d %d", &N, &F);
	int i, radius[N];
	double volume[N];
	double totalVolume = 0;
	for (i = 0; i < N; i++){
		scanf("%d", &radius[i]);
		volume[i] = PI*radius[i]*radius[i]; 
		totalVolume += volume[i]; 
	}
	double maxVolume = totalVolume/(F+1);
	double left = 0, right = maxVolume, mid;
	while (right-left >= EPS){
		mid = left+(right-left)/2;
		int k = 0;
		for (i = 0; i < N; i++){
			k += volume[i]/mid;
		}
		if (k >= F+1){
			maxVolume = mid;
			left = mid+EPS;
		}else{
			right = mid-EPS;
		}
	}
	printf("%.3f\n", maxVolume);
	return 0;
}

002:月度開銷

#include <iostream>
#include <cstdio>

#define PI 3.141592653589793
#define EPS 0.00001 

using namespace std;

int main ()
{	
	int N, M;
	scanf("%d %d", &N, &M);
	int i, cost[N], sum = 0, min, max = 0;
	for (i = 0; i < N; i++){
		scanf("%d", &cost[i]);
		sum += cost[i];
		if (cost[i] > max){
			max = cost[i];
		}
	}
	min = sum;
	int left = max, right = sum, mid;
	while (left <= right){
		mid = left+(right-left)/2;
		int tmp = 0, numberOfFajomonth = 1;
		for (i = 0; i < N; i++){
			tmp += cost[i];
			if (tmp > mid){
				tmp = cost[i];
				numberOfFajomonth++;
			}else if (tmp == mid){
				tmp = 0;
				numberOfFajomonth++;
			}
		}
		if (numberOfFajomonth > M){
			left = mid+1;
		}else{
			if (mid < min){
				min = mid;
			}
			right = mid-1;
		}
	}
	printf("%d\n", min);
	return 0;
}

003:Aggressive cows

#include <iostream>
#include <cstdio>
#include <algorithm>

#define MAXLENGTH 1000000000

using namespace std;

int main ()
{	
	int N, C;
	scanf("%d %d", &N, &C);
	int i, stall[N]; 
	for (i = 0; i < N; i++){
		scanf("%d", &stall[i]);
	}
	sort(stall, stall+N);
	int left = 1, right = MAXLENGTH/C, mid;
	int max = 1;
	while (left <= right){
		mid = left+(right-left)/2;
		int j, k = 0;
		for (j = C; j-1 && k < N; j--){
			int t = stall[k]+mid;
			k++;
			while (k < N && stall[k] < t){
				k++;
			}
		}
		if (j > 1){
			right = mid-1;
		}else{
			if (k >= N){
				right = mid-1;
			}else{
				if (mid > max){
					max = mid;
				}
				left = mid+1;
			}
		}
	}
	printf("%d\n", max);
	return 0;
}

程序設計與算法（一）C語言程序設計第十二週習題(2019夏季)

001:sort簡單題

a+1, a+7, greater<int>()

002:仍是sort簡單題

struct Rule1{
	bool operator()(const int & a1, const int & a2){
		if (a1%10 != a2%10){
			return a1%10 < a2%10;
		}else{
			return a1>a2;
		}
	}
};

struct Rule2{
	bool operator()(const Point & a1, const Point & a2){
		int t1, t2;
		t1 = a1.x*a1.x+a1.y*a1.y;
		t2 = a2.x*a2.x+a2.y*a2.y;
		if (t1 != t2){
			return t2>t1;
		}else{
			if (a1.x != a2.x){
				return a2.x>a1.x;
			}else{
				return a2.y>a1.y;
			}
		}
	}
};

003:Set

#include <iostream>
#include <set>
#include <cstring> 

using namespace std;

struct od{
	char a[4];
	int b;
};

int main()
{
	multiset<int> st;
	multiset<int> st_cpy;
	int n;
	scanf("%d", &n);
	int i;
	struct od order[n];
	for (i = 0; i < n; i++){
		scanf("%s %d", order[i].a, &order[i].b);
	}
	for (i = 0; i < n; i++){
		if (strcmp(order[i].a, "add") == 0){
			st.insert(order[i].b);
			st_cpy.insert(order[i].b);
			multiset<int>::iterator j;
			int t = 0;
			for (j = st.begin(); j != st.end(); j++){
				if (*j == order[i].b){
					t++;
				}
			}
			printf("%d\n", t);
		}
		if (strcmp(order[i].a, "del") == 0){
			multiset<int>::iterator cs;
			int t = 0;
			for (cs = st.begin(); cs != st.end(); cs++){
				if (*cs == order[i].b){
					t++;
				}
			}
			st.erase(order[i].b); 
			printf("%d\n", t);
		}
		if (strcmp(order[i].a, "ask") == 0){
			int ever;
			multiset<int>::iterator k;
			k = st_cpy.find(order[i].b);
			if (k == st_cpy.end()){
				ever = 0;
			}else{
				ever = 1;
			}
			printf("%d ", ever);
			multiset<int>::iterator j;
			int t = 0;
			for (j = st.begin(); j != st.end(); j++){
				if (*j == order[i].b){
					t++;
				}
			}
			printf("%d\n", t);
		}
	}
	return 0;
}

004:熱血格鬥場

#include <iostream>
#include <set>
#include <cstring>
#include <cstdlib>

using namespace std;

struct member{
	int ID;
	long long power;
};

struct rule{
	bool operator()(const struct member & a1, const struct member a2){
		return a1.power < a2.power;
	}
};

struct competition{
	int freshmanID;
	int membershID;
}; 

struct member minp(struct member a, struct member b, struct member c);

int main()
{
	set<struct member, rule> st;
	int n;
	scanf("%d", &n);
	struct member a[n+1], tmpA, tmpB, tmp;
	struct competition c[n];
	a[0].ID = 1;
	a[0].power = 1000000000;
	st.insert(a[0]);
	int i;
	set<struct member, rule>::iterator j; 
	for (i = 1; i < n+1; i++){
		scanf("%d %lld", &a[i].ID, &a[i].power);
		j = st.lower_bound(a[i]);
		tmpA.power = j->power;
		tmpA.ID = j->ID;
		tmpB = tmpA;
		if (j != st.begin()){
			j--;
			tmpB.power = j->power;
			tmpB.ID = j->ID;
		}
		tmp = minp(tmpA, tmpB, a[i]);
		c[i-1].freshmanID = a[i].ID;
		c[i-1].membershID = tmp.ID;
		st.insert(a[i]);
	}
	for (i = 0; i < n; i++){
		printf("%d %d\n", c[i].freshmanID, c[i].membershID);
	}
	return 0;
}

struct member minp(struct member a, struct member b, struct member c){
	int t1, t2;
	t1 = abs(c.power-a.power);
	t2 = abs(c.power-b.power);
	if (t1 >= t2){
		return b;
	}else{
		return a;
	}
}

程序設計與算法（一）C語言程序設計第十三週習題(2019夏季)

001:冷血格鬥場

#include <iostream>
#include <map>
#include <cstring>
#include <cstdlib>

using namespace std;

struct competition{
	int freshmanID;
	int memberID;
};

int main()
{
	map<long long, int> st;
	int n;
	scanf("%d", &n);
	st[1000000000] = 1;
	struct competition c[n];
	int i, id;
	long long power, diff1, diff2;
	map<long long, int>::iterator j, jtmp; 
	for (i = 1; i < n+1; i++){
		scanf("%d %lld", &id, &power);
		j = st.lower_bound(power);
		if (j == st.end()){
			j--;
		}
		diff1 = abs(j->first-power);
		jtmp = j;
		diff2 = diff1;
		if (j != st.begin()){
			j--;
			diff2 = abs(j->first-power);
		}
		if (diff1 > diff2 || diff1 == diff2 && jtmp->second > j->second){
			jtmp = j;
		}
		c[i-1].freshmanID = id;
		c[i-1].memberID = jtmp->second;
		if (jtmp->first == power && jtmp->second < id){
			;
		}else{
			st[power] = id;
		}
	}
	for (i = 0; i < n; i++){
		printf("%d %d\n", c[i].freshmanID, c[i].memberID);
	}
	return 0;
}

程序設計與算法（一）C語言程序設計期末考試(2019夏季)

01:分段函數

#include <iostream>
#include <cstdio>

#define EPS 0.000001

using namespace std;

int main(){
	double x, y;
	scanf("%lf", &x);
	if (x>-EPS && x-5 < -EPS){
		y = -x+2.5;
	}else if (x-5 > -EPS && x-10 < -EPS){
		y = 2-1.5*(x-3)*(x-3);
	}else if (x-10 > -EPS && x-20 < -EPS){
		y = x/2-1.5;
	}
	printf("%.3f", y);
	return 0;
}

02:求分數序列和

#include <iostream>
#include <cstdio>

using namespace std;

int main(){
	double sum = 0;
	int n, p = 1, q = 2, tmp;
	scanf("%d", &n);
	for (int i = 0; i < n; i++){
		sum += 1.0*q/p;
		tmp = q;
		q = q+p;
		p = tmp;
	}
	printf("%.4f", sum);
	return 0;
}

03:畫矩形

#include <iostream>
#include <cstdio>

using namespace std;

int main(){
	int height, width, isFilled;
	char fc;
	scanf("%d %d %c %d", &height, &width, &fc, &isFilled);
	for (int i = 0; i < height; i++){
		if (isFilled){
			for (int j = 0; j < width; j++){
				printf("%c", fc);
			}
		}else{
			if (i == 0 || i == height-1){
				for (int j = 0; j < width; j++){
					printf("%c", fc);
				}
			}else{
				for (int j = 0; j < width; j++){
					if (j == 0 || j == width-1){
						printf("%c", fc);
					}else{
						printf(" ");
					}
				}
			} 
		}
		printf("\n");
	}
	return 0;
}

04:求出e的值

#include <iostream>
#include <cstdio>

using namespace std;

int main(){
	double e = 1;
	int n;
	long long factorial;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++){
		factorial = 1;
		for (int j = 1; j <= i; j++){
			factorial *= j;
		}
		e += 1.0/factorial;
	}
	printf("%.10f", e);
	return 0;
}

05:計算多項式的值

#include <iostream>
#include <cstdio>

using namespace std;

int main(){
	float x, polynomial = 1, product = 1;
	int n;
	scanf("%f %d", &x, &n);
	for (int i = 1; i <= n; i++){
		product *= x;
		polynomial += product;
	}
	printf("%.2f", polynomial);
	return 0;
}

06:大整數加法

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

char a[210], b[210], c[210];

void reverse(char *s);
void plusHugeInt(char *a, char *b, char *sum);
void printHugeInt(char *s);

int main(){
	cin >> a >> b;
	reverse(a);
	reverse(b);
	plusHugeInt(a, b, c);
	printHugeInt(c); 
	return 0;
}

void reverse(char *s){
	int len = strlen(s), tmp;
	int i = 0, j = len-1;
	while (i < j){
		tmp = s[i];
		s[i] = s[j];
		s[j] = tmp;
		i++;
		j--;
	}
	return;
}

void plusHugeInt(char *a, char *b, char *c){
	int lenA = strlen(a), lenB = strlen(b);
	int i, jw = 0, sum;
	for (i = 0; i < lenA && i < lenB; i++){
		sum = a[i]-'0'+b[i]-'0'+jw;
		c[i] = sum%10+'0';
		jw = sum/10;
	}
	while (i < lenA){
		sum = a[i]-'0'+jw;
		c[i] = sum%10+'0';
		jw = sum/10;
		i++;
	}
	while (i < lenB){
		sum = b[i]-'0'+jw;
		c[i] = sum%10+'0';
		jw = sum/10;
		i++;
	}
	if (jw == 1){
		c[i++] = '1';
	}
	c[i] = '\0';
	return;
}

void printHugeInt(char *s){
	int j = -1, lenS = strlen(s);
	for (int i = 0; i < lenS; i++){
		if (s[i]-'0' > 0){
			j = i;
		}
	}
	if (j == -1 ){
		printf("0");
	}else{
		s[j+1] = '\0';
		for (int k = strlen(s)-1; k >= 0; k--){
			printf("%c", s[k]);
		}
	}
	return;
}

07:有趣的跳躍

#include <iostream>
#include <cstdlib>

using namespace std;

int main(){
	int n;
	cin >> n;
	if (n == 1){
		printf("Jolly");
	}else{
		int a[n], difference[n-1], check[n-1], isJolly = 1;
		for (int i = 0; i < n; i++){
			cin >> a[i];
		}
		for (int i = 0; i < n-1; i++){
			check[i] = 0;
			difference[i] = abs(a[i]-a[i+1]);
		}
		for (int i = 0; i < n-1; i++){
			if (difference[i] == 0 || difference[i] > n-1){
				break;
			}else{
				check[difference[i]-1] = 1;
			}
		}
		for (int i = 0; i < n-1; i++){
			if (check[i] == 0){
				printf("Not jolly");
				isJolly = 0;
				break;
			}
		}
		if (isJolly){
			printf("Jolly");
		}
	}
	return 0;
}

08:單詞倒排

#include <iostream>
#include <cstring>

#define MAXSIZE 100

using namespace std;

char s[MAXSIZE];
char *p[MAXSIZE];

int main(){
	for (int i = 0; i < MAXSIZE; i++){
		p[i] = NULL;
	}
	cin.getline(s, MAXSIZE);
	char *tmp;
	tmp = strtok(s, " ");
	int i = 0;
	while (tmp){
		p[i++] = tmp;
		tmp = strtok(NULL, " ");
	}
	for (int j = i-1; j >= 0; j--){
		cout << p[j] << " ";
	}
	return 0;
}

09:行程長度編碼

#include <iostream>
#include <cstring>
#include <cctype>

#define MAXSIZE 1000

using namespace std;

struct charAndNumber{
	char c;
	int n;
};

char s1[MAXSIZE];
struct charAndNumber s2[MAXSIZE];

int main(){
	cin >> s1;
	for (int i = 0; i < MAXSIZE; i++){
		s2[i].c = '\0';
		s2[i].n = 0;
	}
	s2[0].c = toupper(s1[0]);
	s2[0].n++;
	int len = strlen(s1), j = 0;
	for (int i = 1; i < len; i++){
		if (s1[i] == s2[j].c || s1[i]+'A'-'a' == s2[j].c){
			s2[j].n++;
		}else{
			j++;
			s2[j].c = toupper(s1[i]);
			s2[j].n++;
		}
	}
	for (int i = 0; i <= j; i++){
		cout << "(" << s2[i].c << "," << s2[i].n << ")";
	}
	return 0;
}

10:反反覆覆

#include <iostream>
#include <cstring>

#define MAXSIZE 200

using namespace std;

char s[MAXSIZE];

int main(){
	int line, row;
	cin >> row >> s;
	line = strlen(s)/row;
	for (int i = 0; i < row; i++){
		int k = i;
		/* 注意題目中並非把字符串從上到下地一列一列填成的矩陣按每行
		從左到右輸出，而是一行從左到右，一行從右到左。因此獲得的加密後
		的字符串，能夠把從右到左輸出的部分先顛倒，再把一個矩陣一行一行
		輸出。若是不想作顛倒這一步，就能夠像下面這樣，一行輸出整數第a個，
		下一行輸出倒數第a個。 */
		for (int j = 0; j < line; j++){
			printf("%c", s[j*row+k]);
			k = row-1-k;
		}
	}
	return 0;
}

11:網線主管

#include <cstdio>

using namespace std;

/* 這題一開始用的double，怎麼都不行，有的時候本身電腦上能夠，傳上去也不行，
實在是沒法理解，不知道怎麼用double的變量來作。我是隻能統統乘100用整數來了 */
int main(){
	int N, K;
	scanf("%d %d", &N, &K);
	int a[N];
	/* 注意，單位用釐米時，全部的都乘100，和sum用8字節的int可能就不夠了。
	我有個測試數據過不了就是由於這個。 */
	long long sum = 0;
	double tmpA;
	for (int i = 0; i < N; i++){
		scanf("%lf", &tmpA);
		a[i] = tmpA*100;
		sum += a[i];
	}
	/* 枚舉+二分法 */
	/* 注意，left不能用0，否則下面a[i]/mid的時候，除數可能爲0，就出錯了，
	有個測試數據過不了就是由於一開始left是0。我電腦上的提示是
	float exception:8。反正就是出錯。何況題目中要求若是小於max小於1就輸出0.00，
	因此也left也不必設置成1如下的。 */
	int left = 1, right = sum/K, mid, max = 0;
	int tmp;
	while (left <= right){
		mid = left+(right-left)/2;
		tmp = 0;
		for (int i = 0; i < N; i++){
			tmp += a[i]/mid;
		}
		if (tmp >= K){
			max = mid;
			left = mid+1;
		}else{
			right = mid-1;
		}
	}
	if (max < 1){
		printf("0.00");
	}else{
		printf("%.2f", 1.0*max/100);
	}
	return 0;
}

12:查找最接近的元素

#include <iostream>

using namespace std;

int main(){
	int n, m;
	cin >> n;
	int a[n];
	for (int i = 0; i < n; i++){
		cin >> a[i];
	}
	cin >> m;
	int test, left, right, mid, l, r;
	for (int i = 0; i < m; i++){
		cin >> test;
		left = 0;
		right = n-1;
		l = a[0];
		r = a[n-1];
		while (left <= right){
			mid = left+(right-left)/2;
			if (a[mid] > test){
				r = a[mid];
				right = mid-1;
			}else{
				l = a[mid];
				left = mid+1;
			}
		}
		if (l == test){
			printf("%d\n", l);
		}else{
			if (test-l <= r-test){
				printf("%d\n", l);
			}else{
				printf("%d\n", r);
			}
		}
	}
	return 0;
}

13:1的個數

#include <iostream>

using namespace std;

int main(){
	int N;
	cin >> N;
	int n = N, cnt = 0;
	for (int i = 0; i < 32; i++){
		if (n&1 == 1){
			cnt++;
		}
		n >>= 1;
	}
	printf("%d", cnt);
	return 0;
}

14:判斷元素是否存在

#include <iostream>
#include <cstdio>

#define MAXSIZE 100000

using namespace std;

int main(){
	long long a[MAXSIZE];
	int k, x;
	int isFind = 0;
	scanf("%d,%d", &k, &x);
	a[0] = k;
	for (int i = 1; i < MAXSIZE; i++){
		if (i%2 == 1){
			a[i] = 2*a[(i+1)/2-1]+1;
		}else{
			a[i] = 3*a[(i+1)/2-1]+1;
		}
	}
	for (int i = 0; i < MAXSIZE; i++){
		if (a[i] == x){
			isFind = 1;
			break;
		}
	}
	if (isFind){
		printf("YES");
	}else{
		printf("NO");
	}
	return 0;
}