acdream 1726 A Math game (部分和问题 DFS剪枝)_部分和问题加剪枝-优快云博客

本文介绍了一道名为AMathgame的问题，该问题是判断是否能从一组整数中选取若干个数，使得这些数的总和等于给定的目标值H。文章提供了使用DFS算法结合前缀和剪枝的方法来解决此问题，并附带了完整的C++实现代码。

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A Math game

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

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 areintegers.

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

Source

第九届北京化工大学程序设计竞赛

题目链接： http://acdream.info/problem?pid=1726

题目大意：部分和问题

题目分析：正解是二分＋DFS，我的做法是DFS用前缀和剪枝，4ms过，估计还是数据水了，

#include <cstdio>
#include <cstring>
#define ll long long
ll a[45], h, sum[45];
int n;
bool flag;
 
void DFS(ll num, int pos)
{
    if(num == h)
    {
        flag = true;
        return;
    }
    if(flag || pos == n + 1)  
        return;
    //若当前数字加剩下的数字和小于h或者当前数字大于h则返回
    //不加妥T
    if(sum[n] - sum[pos - 1] + num < h || num > h) 
        return;
    DFS(num + a[pos], pos + 1);
    DFS(num, pos + 1);
    return;
}
 
int main()
{
    while(scanf("%d %lld", &n, &h) != EOF)
    {
        memset(sum, 0, sizeof(sum));
        flag = false;
        for(int i = 1; i <= n; i++)
        {
            scanf("%lld", &a[i]);
            sum[i] = sum[i - 1] + a[i];
        }
        DFS(0, 1);
        if(flag)
            printf("Yes\n");
        else
            printf("No\n");
    }
}