HDU 5665

Lucky

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 138    Accepted Submission(s): 96

Problem Description

     Chaos August likes to study the lucky numbers.

     For a set of numbers S,we set the minimum non-negative integer,which can't be gotten by adding the number in S,as the lucky number.Of course,each number can be used many times.

     Now, given a set of number S, you should answer whether S has a lucky number."NO" should be outputted only when it does have a lucky number.Otherwise,output "YES".

 

Input

     The first line is a number T,which is case number.

     In each case,the first line is a number n,which is the size of the number set.

     Next are n numbers,means the number in the number set.

    1≤n≤105,1≤T≤10,0≤ai≤109 .

 

Output

     Output“YES”or “NO”to every query.

 

Sample Input

1

1

2

 

Sample Output

NO

 

Source

BestCoder Round #80

 

题意：集合S 中 有若干数 如果能够通过任意 组合（不限次数）加和得到所有 自然数 则输出“YES” 否则“NO”

 

 

题解：自然数包含0 

        判断可以发现 只有当S中存在 “0”与“1”时才会满足条件输出“YES”

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#define ll __int64
#define pi acos(-1.0)
using namespace std;
int t;
int n;
ll a;
int main()
{
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        scanf("%d",&n);
        int flag1=0;
        int flag2=0;
        for(int j=1;j<=n;j++)
        {
            scanf("%I64d",&a);
            if(a==1)
            {
            flag1=1;
            }
            if(a==0)
            {
            flag2=1;
            }
        }
        if(flag1==1&&flag2==1)
         cout<<"YES"<<endl;
         else
         cout<<"NO"<<endl;
    }
    
    return 0;
}