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=;i<=t;i++)
     {
         scanf("%d",&n);
         int flag1=;
         int flag2=;
         for(int j=;j<=n;j++)
         {
             scanf("%I64d",&a);
             if(a==)
             {
             flag1=;
             }
             if(a==)
             {
             flag2=;
             }
         }
         if(flag1==&&flag2==)
          cout<<"YES"<<endl;
          else
          cout<<"NO"<<endl;
     }

     return ;
 }