The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple

Lucky 7

Time Limit: 1 Second Memory Limit: 65536 KB

BaoBao has just found a positive integer sequence of length from his left pocket and another positive integer from his right pocket. As number 7 is BaoBao's favorite number, he considers a positive integer lucky if is divisible by 7. He now wants to select an integer from the sequence such that is lucky. Please tell him if it is possible.

Input

There are multiple test cases. The first line of the input is an integer (about 100), indicating the number of test cases. For each test case:

The first line contains two integers and (), indicating the length of the sequence and the positive integer in BaoBao's right pocket.

The second line contains positive integers (), indicating the sequence.

Output

For each test case output one line. If there exists an integer such that and is lucky, output "Yes" (without quotes), otherwise output "No" (without quotes).

Sample Input

Sample Output

No

Yes

Yes

Yes

Hint

For the first sample test case, as 4 + 7 = 11, 5 + 7 = 12 and 6 + 7 = 13 are all not divisible by 7, the answer is "No".

For the second sample test case, BaoBao can select a 7 from the sequence to get 7 + 7 = 14. As 14 is divisible by 7, the answer is "Yes".

For the third sample test case, BaoBao can select a 5 from the sequence to get 5 + 2 = 7. As 7 is divisible by 7, the answer is "Yes".

For the fourth sample test case, BaoBao can select a 100 from the sequence to get 100 + 26 = 126. As 126 is divisible by 7, the answer is "Yes".

原题地址：http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5762

题意：

给你一个n，和b，然后一个长度为n的数组A 问你是否有数组A的元素加上B能被7整除如果可以输出Yes，不能输出No

代码：

#include<bits/stdc++.h>

using namespace std;

int a[];

int main()

{

    std::ios::sync_with_stdio(false);

    int t;

    while(cin>>t){

        while(t--){

        int n,num;

        cin>>n>>num;

        int flag=;

        for(int i=;i<n;i++){

                cin>>a[i];

        }

        for(int i=;i<n;i++){

                if((a[i]+num)%==){

                        flag=;break;

                }

        }

        if(flag)cout<<"Yes"<<endl;

        else cout<<"No"<<endl;

    }

    }

    return ;

}

巴特西

The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple - M Lucky 7

Input

Output

Sample Input

Sample Output

Hint

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