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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
,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
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
#include
#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"); } }
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