Description:

BaoBao has just found a positive integer sequence a 1 , a 2 , … … , a n a_{1},a_{2},……,a_{n} a1​,a2​,……,an​ of length n from his left pocket and another positive integer b 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 a k a_{k} ak​ from the sequence such that ( a k + b a_{k} + b ak​+b ) is lucky. Please tell him if it is possible.

Input:

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

The first line contains two integers n and b ( 1 ≤ n , b ≤ 100 1 \le n,b \le 100 1≤n,b≤100 ), indicating the length of the sequence and the positive integer in BaoBao’s right pocket.

The second line contains n positive integers a 1 , a 2 , … … , a n a_{1},a_{2},……,a_{n} a1​,a2​,……,an​ ( 1 ≤ n ≤ 100 1 \le n \le 100 1≤n≤100 ), indicating the sequence.

Output:

For each test case output one line. If there exists an integer a k a_{k} ak​ such that a k ∈ { a 1 , a 2 , … … , a n } a_{k} \in \{a_{1},a_{2},……,a_{n}\} ak​∈{a1​,a2​,……,an​} and ( a k + b ) (a_{k} + b) (ak​+b) is lucky, output “Yes” (without quotes), otherwise output “No” (without quotes).

Sample Input:

4
3 7
4 5 6
3 7
4 7 6
5 2
2 5 2 5 2
4 26
100 1 2 4

Sample Output:

No
Yes
Yes
Yes

题目连接

直接枚举判断是否存在 ( a i + b ) % 7 = = 0 (a_{i}+b) \% 7 == 0 (ai​+b)%7==0 。

AC代码:

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <iomanip>
#include <cctype>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdlib>
#include <sstream>
#include <set>
#include <map>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const int maxn = 1e2+5;
const double eps = 1e-5;
const double pi = asin(1.0)*2;
const double e = 2.718281828459;

int t;
int n, b;
int a;
bool flag;

int main() {
    //fropen("in.txt", "r", stdin);
    scanf("%d", &t);
    while (t--) {
        flag = 0;
        scanf("%d %d", &n, &b);
        for (int i = 0; i < n; ++i) {
            scanf("%d", &a);
            if (flag) {
                continue;
            }
            else {
                if ((a + b) % 7 == 0) {
                    flag = 1;
                }
            }
        }
        if (!flag) {
            printf("No\n");
        }
        else {
            printf("Yes\n");
        }
    }
    return 0;
}