歐拉計劃16-20題

時間 2019-11-09

標籤歐拉計劃欄目快樂工作简体版

原文原文鏈接

1六、Power digit sumphp

2¹⁵ = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.git

What is the sum of the digits of the number 2¹⁰⁰⁰?web

題目大意:數組

2¹⁵ = 32768 而且其各位之和爲 is 3 + 2 + 7 + 6 + 8 = 26.less

2¹⁰⁰⁰ 的各位數之和是多少？ide

#include <stdio.h> 
#include <stdbool.h>

void solve(void)
{
    int a[100000] = {0};
    int n, sum, i, j;
    n = sum = 0;
    a[0] = 1;
    for(i = 0; i < 1000; i++) {  //以1000進制的方法存儲
        for(j = 0; j <= n; j++) {
            a[j] *= 2;
        }
        for(j = 0; j <= n; j++) {
            if(a[j] >= 10000) {
                a[j] %= 10000;
                a[j+1]++;
                n++;
            }
        }
    }
    for(i = 0; i <= n; i++) {
        sum += a[i] / 10000;
        a[i] %= 10000;
        sum += a[i] / 1000;
        a[i] %= 1000;
        sum += a[i] / 100;
        a[i] %= 100;
        sum += a[i] / 10;
        a[i] %= 10;
        sum += a[i];

    }
    printf("%d\n",sum);
}

int main(void)
{
    solve();
    return 0;
}

View Code

Answer:1366C

ompleted on Sun, 17 Nov 2013, 15:23

1七、Number letter countsui

If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.this

If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?spa

NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage..net

題目大意：

若是用英文寫出數字1到5: one, two, three, four, five, 那麼一共須要3 + 3 + 5 + 4 + 4 = 19個字母。

若是數字1到1000（包含1000）用英文寫出，那麼一共須要多少個字母？

注意: 空格和連字符不算在內。例如，342 (three hundred and forty-two)包含23個字母； 115 (one hundred and fifteen)包含20個字母。"and" 的使用與英國標準一致。

#include <stdio.h> 
#include <stdbool.h>

int a[101] = {0,3,3,5,4,4,3,5,5,4,3,6,6,8,8,7,7,9,8,8};

void init(void)  //初始化數組
{
    a[20] = 6;
    a[30] = 6;
    a[40] = 5;
    a[50] = 5;
    a[60] = 5;
    a[70] = 7;
    a[80] = 6;
    a[90] = 6;
    a[100] = 7;
}

int within100(void)  //計算1~99所含字母的和
{
    int i, sum, t;
    t = sum = 0;
    for(i = 1; i <= 9; i++) t += a[i];
    for(i = 1; i <= 19; i++) sum += a[i];
    for(i = 2; i <= 9; i++) {
        sum += a[i*10] * 10;
        sum += t;
    }
    return sum;
}

void solve(void)
{
    int i;
    int sum, t;
    sum = t = within100();
    for(i = 1; i < 10; i++) {
        sum += (a[i] + 10) * 99 + (a[i] + 7) + t;
    }
    sum += 11;

    printf("%d\n",sum);
}

int main(void)
{
    init();
    solve();
    return 0;
}

View Code

Answer:21124

Completed on Sun, 17 Nov 2013, 16:30

1八、Maximum path sum I

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

7 4

2 4 6

8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

題目大意：

找出從如下三角形的頂端走到底端的最大總和：

#include<stdio.h>

#define N 15
int main()
{
    char t[5];
    int s[N][N]={0};
    FILE *f;
    int i,j;
    f = fopen("18.txt","r");
    for (i = 0; i < N; i++) {
        for (j = 0; j <= i; j++) {
            fgets(t,4,f);
            s[i][j] =atoi(t);
        }
    }
    fclose(f);
    for ( i = N-2; i >=0; i--) {
          for ( j = 0; j <= i; j++) {
               if (s[i+1][j] > s[i+1][j+1]) {
                s[i][j]+=s[i+1][j];
            } else {
                s[i][j]+=s[i+1][j+1];
            }
        }
    }
    printf("answer: %d\n",s[0][0]);
    return 0;
}

View Code

Answer:1074

Completed on Thu, 1 May 2014, 16:31

1九、Counting Sundays

You are given the following information, but you may prefer to do some research for yourself.

1 Jan 1900 was a Monday.
Thirty days has September, April, June and November. All the rest have thirty-one, Saving February alone, Which has twenty-eight, rain or shine. And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

題目大意：

如下是一些已知信息，可是或許你須要本身作一些其餘的調查。

1900年1月1日是星期一。
30天的月份有：9月，4月，6月，11月。
此外的月份都是31天，固然2月除外。
2月在閏年有29天，其餘時候有28天。
年份能夠被4整除的時候是閏年，可是不能被400整除的世紀年（100的整數倍年）除外。

20世紀（1901年1月1日到2000年12月31日）一共有多少個星期日落在了當月的第一天？

#include <stdio.h> 
#include <stdbool.h>

const int a[2][12] = {{31,28,31,30,31,30,31,31,30,31,30,31},
                     {31,29,31,30,31,30,31,31,30,31,30,31}};


bool leapYear(int n)  //判斷閏年
{
    return (((n % 4 ==0) && (n % 100 !=0)) || (n % 400 == 0));
}

bool issunday(int n) //判斷某天是不是星期天
{
    return (n % 7 == 0 ? true : false);
}

void solve(void)
{
    int num, i, j, count;
    count = 0;

    i = 1901;
    num = 1;
    while(i < 2000) {

        int t = (leapYear(i) ? 1 : 0);   //判斷閏年
        for(j = 0; j < 12; j++) {
            num += a[t][j];
            if(issunday(num)) count++;
        }
        i++;
    }
    printf("%d\n",count);
}

int main(void)
{
    solve();
    return 0;
}

View Code

Answer:171

Completed on Mon, 18 Nov 2013, 03:38

20、Factorial digit sum

n! means n (n 1) ... 3 2 1

For example, 10! = 10 9 ... 3 2 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

Find the sum of the digits in the number 100!

題目大意：

n! = n (n 1) ... 3 2 1

例如， 10! = 10 9 ... 3 2 1 = 3628800, 那麼10!的各位之和就是3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

算出100!的各位之和。

#include <stdio.h>
#include <math.h>

#define N 100

int main(void){
    int n=N*log(N/3),a[n],ca=0,i,j;
    for(i = 0; i < n; i++)
        a[i] = 0;
    a[n-1] = 1;
    for(i = 1; i <= N; i++){
        for(j = n - 1; j >= 0; j--){
            ca = i * a[j] + ca;
            a[j] = ca % 10;
            ca /= 10;
        }
        ca = 0;
    }
    ca = 0;
    for(i = 0; i < n; i++)
        ca += a[i];
    printf("%d\n",ca);
    return 0;
}