HDU 5665 Lucky （水题）

Lucky

题目链接：

http://acm.hust.edu.cn/vjudge/contest/123316#problem/G

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.

.

Output

Output“YES”or “NO”to every query.

Sample Input

1
1
2

Sample Output

No

题意：

给出一个具有n个数字的集合；
定义lucky number：不能由集合中的数字相加得到的最小的数.

题解：

能否构成任意正数取决于是否含有1，有1则一定能，否则一定不能.
注意题目说的是非负整数，则还要考虑0; 只有当含有0的时候才能得到0.

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define eps 1e-8
#define maxn 1100
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n;

int main(int argc, char const *argv[])
{
    //IN;

    int t;  scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);

        int flag = 0;
        int flag2 = 0;
        for(int i=1; i<=n; i++) {
            int x; scanf("%d", &x);
            if(x == 1) flag = 1;
            if(x == 0) flag2 = 1;
        }

        if(flag && flag2) printf("YES
");
        else printf("NO
");
    }

    return 0;
}