acdream A - A Math game

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#acdream #A Math game

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本文介绍了一种通过深度优先搜索解决特定数学游戏的算法。游戏中玩家需从一组数中挑选若干数，使其总和等于给定的目标值。文章提供了一个示例程序，使用递归实现的深度优先搜索来判断是否存在一种选择方案使总和达到目标值。

A - A Math game

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

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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

dfs，每个数字根据选与不选深搜，当当前和大于H或当前和加上剩余和小于H直接跳出。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

int n;
long long m;
long long num[100];
long long sum[100];
bool f;
void dfs(int x,long long s)
{
if (s==m){
  f=true;return;
}
if (f) return;
if (s>m||x==n) return;
if (sum[n-1]-sum[x]+s+num[x]<m)
  return;
dfs(x+1,s+num[x]);
dfs(x+1,s);
}
int main()
{
int i,j;
while (scanf("%d%lld",&n,&m)!=EOF)
{
  for (i=0;i<n;i++)
   scanf("%lld",&num[i]);
  memset(sum,0,sizeof(sum));
  sum[0]=num[0];
  for (i=1;i<n;i++)
   sum[i]=sum[i-1]+num[i];
  f=false;
  dfs(0,0);
  if (!f) printf("No\n");
  else printf("Yes\n");
}

}