ACdream1726 A Math game(dfs)

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本文介绍了一个名为AMathgame的编程题目，该题目要求玩家从一组数字中选择若干数，使得这些数的总和等于给定的目标数H。文章提供了一种深度优先搜索（DFS）算法解决方案，并给出了完整的AC代码示例。

A Math game

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 256000/128000KB (Java/Others)

Submit Statistic Next Problem

Problem Description

Recently, Losanto find an interesting Math game. The rule is simple: Tell you a number H, and you can choose some numbers from a set {a[1],a[2],......,a[n]}.If the sum of the number you choose is H, then you win. Losanto just want to know whether he can win the game.

Input

There are several cases.
In each case, there are two numbers in the first line n (the size of the set) and H. The second line has n numbers {a[1],a[2],......,a[n]}. 0<n<=40, 0<=H<10^9, 0<=a[i]<10^9,All the numbers are integers.

Output

If Losanto could win the game, output "Yes" in a line. Else output "No" in a line.

Sample Input

10 87
2 3 4 5 7 9 10 11 12 13
10 38
2 3 4 5 7 9 10 11 12 13

Sample Output

No
Yes

简单的dfs，从大到小dfs即可。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
using namespace std;
const int MAXN = 45;
int n, h, a[MAXN], sum[MAXN];
bool flag;
void dfs(int n, int s)
{
	if(flag) return;
	if(s > sum[n]) return;
	if(sum[n] == s || s == 0) {
		flag = true;
		return;
	}
	for(int i = n; i >= 1; --i)
		if(s >= a[i]) dfs(i - 1, s - a[i]);
}
int main(int argc, char const *argv[])
{
	while(scanf("%d%d", &n, &h) != EOF) {
		memset(sum, 0, sizeof(sum));
		flag = false;
		for(int i = 1; i <= n; ++i) {
			scanf("%d", &a[i]);
			sum[i] = sum[i - 1] + a[i];
		}
		dfs(n, h);
		if(flag) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}