HDU (a+k)%7 阅读题

Problem K

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 60   Accepted Submission(s) : 55

Problem Description

BaoBao has just found a positive integer sequence $a_1, a_2, \dots, a_n$ 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 $x$ lucky if $x$ is divisible by 7. He now wants to select an integer $a_k$ from the sequence such that $(a_k+b)$ 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 $b$ ($1 \le n, b \le 100$), indicating the length of the sequence and the positive integer in BaoBao's right pocket.

The second line contains $n$ positive integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), indicating the sequence.

Output

For each test case output one line. If there exists an integer $a_k$ such that $a_k \in \{a_1, a_2, \dots, a_n\}$ and $(a_k + b)$ 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

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".

 给定数列如果存在 （ai+k）%7==0输出Yes否则输出No

#include 
#include 
#include 
#include 
#include 

using namespace std ; 

#define maxn 110000
int num[maxn] ; 

int main(){
    int t ; 
    int n , m ; 

    cin >> t ; 
    while(t--){
        cin >> n >> m ; 
        bool flag = false ; 
        for(int i=1 ; i<=n ; i++){
            cin >> num[i] ; 
            num[i] = num[i] + m ;
            if(num[i]%7==0){
                flag = true ; 
            } 
        }
        if(flag == true){
            cout << "Yes"<<endl ; 
        }else {
            cout<<"No"<<endl ; 
        }
    }

    return 0 ; 
}

 

转载于:https://www.cnblogs.com/yi-ye-zhi-qiu/p/9064954.html