杭电OJ第十五届ACM第一题 Hearthstone

  Cdfpysw loves playing a card game called "Hearthstone".
  Now he has N cards, he wants to split these cards into 4 piles. Let‘s assume the number of cards in each pile is a1, a2, a3, a4.
  It must be satisfied that:
    a1 * k1 = a2 + a3 + a4
    a2 * k2 = a1 + a3 + a4
    a3 * k3 = a1 + a2 + a4
    a1, a2, a3, a4 must be positive
  Because Cdfpysw is clever, there must be a way to split there card. Can you tell Cdfpysw what is the way?

  The first line is an integer T, means the number of cases.
  Then T lines, each line contains 4 integers, N, k1, k2, k3, meaning of these have been shown in the description.
  T <= 100
  1 <= N <= 109
  1 <= k1, k2, k3 <= 200

  For each case, you should output "Case #i: " first, i is the case number.
  Then output 4 integers a1, a2, a3, a4, represent the number of cards in each pile.

1
120 2 3 4

Case #1: 40 30 24 26

import java.util.Scanner;

public class Hearthstone {
	public static int[] a = {1,1,1,1,1};
	//基于深度搜索优先的回溯法
	public static void DFS(int deep,int n,int[] k,int i){
		if(a[1] == n) return;
		
		if(a[deep]==n){
				//当子节点遍历到n,回溯
				a[deep-1]++;
				DFS(deep-1,n,k,i);
		}
		//到第四层，若有解输出并返回，无解则继续向右
		if(deep == 4){
			if((a[1]+a[2]+a[3]+a[4]) == n){
				System.out.println("case #"+i+": "+a[1]+" "+a[2]+" "+a[3]+" "+a[4]);
				return;
			}
			else{
				a[deep]++;
				DFS(deep,n,k,i);
			}
		}
		//满足条件便往下搜索，不满足则剪枝
		if(a[deep] < n && deep < 4){
			if(a[deep]*k[deep-1] == (n-a[deep])){
				DFS(deep+1,n,k,i);
			}
			else{
				a[deep]++;
				DFS(deep,n,k,i);
			}
		}
	}
	public static void main(String[] args) {
		int T;
		int n[] = new int[100];
		int k1[] = new int[100];
		int k2[] = new int[100];
		int k3[] = new int[100];
		int a=1,b=1,c=1,d=1;
		Scanner scanner = new Scanner(System.in);
		T = scanner.nextInt();
		for(int i = 1;i <= T;i++){
			n[i] = scanner.nextInt();
			k1[i] = scanner.nextInt();
			k2[i] = scanner.nextInt();
			k3[i] = scanner.nextInt();
	
		}
		//对每一组实例进行求解
		for(int i = 1;i <= T;i++){
			int[] k = {k1[i],k2[i],k3[i]};
			DFS(1,n[i],k,i);
		}
		//刚开始用的五层大循环。。醉了
		/*for(int i = 1;i <= T;i++){
			for(a = 1;a<=n[i];a++){
				if(a*k1[i] == (n[i]-a))
				for(b = 1;b<=n[i]-a;b++){
					if(b*k2[i] == (n[i]-b))
					for(c = 1;c<=n[i]-a-b;c++){
						if(c*k3[i] == (n[i]-c))
						for(d = 1;d<=n[i]-a-b-c;d++){
							if((a+b+c+d) == n[i]){
								System.out.println("case #"+i+": "+a+" "+b+" "+c+" "+d);
								
							}
						}
					}
				}
			}
		}*/
	}

}

杭电OJ第十五届ACM第一题 Hearthstone

原文：http://blog.csdn.net/lu_chaoqun/article/details/44707883