hdoj 5665 Lucky 【水】

Lucky

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 554    Accepted Submission(s): 327

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

112

 

Sample Output

NO

 

Source

BestCoder Round #80

 

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代码：

//这道题的意思太难理解了！/*    在给定的集合s中，通过相加，不能够得到的最小非负整数为幸运数，如果不存在幸运数，输出YES;否则输出NO； 技巧：如果这个集合里面没有0的话，0肯定不能通过其他的数相加得到，所以0必须有，如果没有1的话，其他的数进行加的话1肯定不能够得到，所以1也必须有，其他的数都可以通过0和多个1相加得到，所以只要这个集合有0和1就能组成所有的非负整数，就不存在幸运数了！ */ #include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int a[100005];int main(){int T;scanf("%d",&T);while(T--){int n;scanf("%d",&n);int flag1=0,flag2=0;for(int i=0;i<n;i++){scanf("%d",&a[i]);if(a[i]==0){flag1=1;}if(a[i]==1){flag2=1;}}if(flag1&&flag2){printf("YES\n");}else{printf("NO\n");}}return 0;}