hdoj 5665 Lucky 【水题】

题目链接：hdoj 5665 Lucky

Lucky

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 465 Accepted Submission(s): 282

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

题意：给定一个集合，你可以使用任意一个元素多次。问你该集合能否得到所有的非负数。

特判0和1即可。

#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <iostream>#include <cmath>#include <queue>#include <stack>#define ll o<<1#define rr o<<1|1#define CLR(a, b) memset(a, (b), sizeof(a))using namespace std;typedef long long LL;typedef pair<int, int> pii;const int INF = 0x3f3f3f3f;const int MAXN = 1e5 + 10;int a[MAXN];int main(){    int t; scanf("%d", &t);    while(t--) {        int n; scanf("%d", &n);        bool flag = false;        bool zero = false;        for(int i = 0; i < n; i++) {            scanf("%d", &a[i]);            if(a[i] == 1) {                flag = true;            }            if(a[i] == 0) {                zero = true;            }        }        printf(flag&&zero ? "YES\n" : "NO\n");    }    return 0;}