第九届北京化工大学程序设计竞赛网络同步赛 A （搜索）

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题解

直接搜索，输入时候记录一下前缀和，然后搜到当前这个数，后面的数加起来都小于需要的话就直接返回

代码

#include <cstdio>#include <queue>#include <cstring>#include <iostream>#include <cstdlib>#include <algorithm>#include <vector>#include <map>#include <string>#include <set>#include <ctime>#include <cmath>#include <cctype>using namespace std;#define maxn 1005#define LL long longint n,m,a[maxn];int sum[maxn];int flag;void dfs(int pos,int y){    if (y==n)        flag=1;    if (flag)        return;    if (y+sum[m]-sum[pos-1] <n || pos>m || y>n)        return;    dfs(pos+1,y+a[pos]);    dfs(pos+1,y);}int cas=1,T;int main(){    while (scanf("%d%d",&m,&n)!=EOF)    {        sum[0]=0;        for (int i = 1;i<=m;i++)        {           scanf("%d",&a[i]);           sum[i]=sum[i-1]+a[i];        }        flag = 0;        dfs(1,0);        if (flag == 0)            printf("No\n");        else            printf("Yes\n");    }    return 0;}

题目

A - 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

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

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