HEU預熱賽

A題：
一道dp的題目
dp[i][j] = k 表明 i行放j個棋子有k中可能
dp[i][j] = dp[i-1][0] + dp[i-1][1] + dp[i-1][2] +...dp[i-1][j]
初始 dp[1][0] = 1, dp[1][1] = 1

注意點：

1. BigInteger 不然爆棧
2. 最後return 從1開始加

import java.util.*;
import java.math.*;

public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        BigInteger[][] dp = new BigInteger[110][110];
        for (int i = 0;i <= 109; i++)
            for(int j = 0;j <= 109; j++)
                dp[i][j] = new BigInteger("0");
        dp[1][0] = new BigInteger("1");
        dp[1][1] = new BigInteger("1");
        
        int n = sc.nextInt();
        for (int i = 2; i <= n ;i++){
            for (int j = 0; j <= i; j++){
                for (int k = 0; k <= j; k++){
                    dp[i][j] = dp[i][j].add(dp[i-1][k]);
                } 
            }
        }
        BigInteger ans = new BigInteger("0");
        for (int i = 1; i <= n; i++)
            ans = ans.add(dp[n][i]);
        System.out.println(ans);
    }
}

B題：用公式
不知道這個A的伴隨矩陣公式，就很難受了- - 。。

import java.util.*;

public class Main2{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            int[][] A = new int[3][3];
            for (int m = 0; m < 3; m ++) {
                for (int n = 0; n < 3; n++) {
                    A[m][n] = sc.nextInt();
                }
            }
            long ao = (A[0][0]* A[1][1]* A[2][2]) + (A[1][0]* A[2][1]* A[0][2]) + (A[2][0]* A[0][1]* A[1][2])
                    - (A[0][2]* A[1][1]* A[2][0]) - (A[0][0]* A[1][2]* A[2][1]) - (A[0][1]* A[1][0]* A[2][2]);
            System.out.println(ao* ao);
        }
    }
}

C題：
一開始理解錯了，他的意思是a^n，我還覺得要a *=a;

import java.math.BigInteger;
import java.util.*;

public class Main3{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            BigInteger a = new BigInteger("0"), n = new BigInteger("0"), b = new BigInteger("0");
            a = sc.nextBigInteger();
            n = sc.nextBigInteger();
            b = sc.nextBigInteger();
            System.out.println(a.modPow(n,b));
        }
    }
}

E題：
感受巨他媽坑了，原本

××if (y1_xing == C)
××B --;
我都服了，被這個卡了小半天。。

import java.math.BigInteger;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.*;

public class Main{
    static int y1_xing; 
    static boolean r;
    static int ny, nm, nd;
    static int[] monthofday = new int[]{-1,31,28,31,30,31,30,31,31,30,31,30,31};
    static int res;     //這個月的幾號
    
    public static boolean is_run(int year) {
        if (year % 400 == 0 ||(year % 4 == 0 && year % 100 != 0))
            return true;
        return false;
    }
    
    public static void gety1_xing(int ny,int y,int A, int B, int C) {
        y1_xing = 1;
        for (ny = 1850; ny < y; ny++) {
            for (nm = 1; nm <= 12; nm ++) {
                int month_ofday = monthofday[nm];
                if(is_run(ny) && nm == 2)   month_ofday += 1;
                for(nd = 1; nd <= month_ofday; nd++) {
                    y1_xing += 1;
                    if (y1_xing == 8)   y1_xing = 1; 
                }
            }
        }
        for(nm = 1;nm < A;nm++) {
            int month_ofday = monthofday[nm];
            if(is_run(y) && nm == 2) month_ofday += 1;
            for(nd = 1; nd <= month_ofday; nd++) {
                y1_xing += 1;
                if (y1_xing == 8)   y1_xing = 1; 
            }
        }
        boolean has = false;
        
        int month_ofday = monthofday[nm];
        if(is_run(y) && B == 2) month_ofday += 1;
        
        for (res = 1;res <= month_ofday; res++) {
            y1_xing += 1;
            if (y1_xing == 8) {
                y1_xing = 1;
            }
            if (y1_xing == C)
                B --;
            
            if (B == 0) {
                has = true;
                break;
            }
            
        }
        if(!has)    res = -1;
    }
    
    public static void main(String[] args) throws ParseException{
        Scanner sc = new Scanner(System.in);
        while (sc.hasNextInt()) {
            int A = sc.nextInt(), B = sc.nextInt(), C = sc.nextInt(), y = sc.nextInt();
            gety1_xing(ny,y, A, B, C);
            if (res != -1) {
                Calendar c = Calendar.getInstance();
                c.set(Calendar.YEAR, y);
                c.set(Calendar.MONTH, A - 1);
                c.set(Calendar.DAY_OF_MONTH, res);
                Date d=new Date();
                d=c.getTime();
                SimpleDateFormat sdf=new SimpleDateFormat("yyyy/MM/dd");
                String str=sdf.format(d);
                System.out.println(str);
            }
            else
                System.out.println("none");
        }
    }
}

H題：
先打表，在找規律！

import java.math.BigInteger;
import java.util.*;

public class MainH{
    //打表 ->  找規律！
    //100 1-1 1-2 1-3 1-4....1-100
    //99 1-1 1-2 1-3...1-99
    //...
    //2 1-1 1-2
    //1 1-1
    public static void main(String[] args){
        /*
        for(int i = 1; i <= 10 ; i++) {
            int[] a = new int[i + 1];   //計數從1開始
            //通過i次
            for(int j = 2; j <= i; j++) {
                for(int k = 1; k*j < a.length; k++) {
                    a[k*j] = 1 - a[k*j];
                }
            }
            for (int j = 1; j <= i; j++) {
                System.out.print(a[j]+" ");
            }
            System.out.println();
        }*/
        
        //找到規律：規律是：
        //通過i次，爲1的位置爲：1 4 9 16 25 36 49 64 81 100 121 144
        //                  A=1,B=3
        //4 - 1 - 1 = 2, 9 - 4 - 1 =4,
        //16 - 9 - 1 = 6, 25 - 16 - 1 = 8
        //依次增加2
        
        //1: 0 0 0 0 0
        //2: 0 1 0 1 0
        //3: 0 1 1 1 0
        //4: 0 1 1 0 0
        //5: 0 1 1 0 1
        
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            long N = sc.nextLong(), A = sc.nextLong(),B = sc.nextLong();
            int start = 1;
            int gap = 2;
            while (start < A) {
                start += gap + 1;
                gap += 2;
            }
            //start >= A
            int res = 0;
            while (start <= B) {
                res ++;
                start += gap + 1;
                gap += 2;
            }
            System.out.println(B - A+1-res);
        }
    }
}

感受沒有練習過ACM 的確是作得磕磕拌拌，全是坑！！！ 明天接着更剩下的題吧- - 。。