ZOJ--Lucky 7

最新推荐文章于 2025-12-10 19:39:18 发布

原创最新推荐文章于 2025-12-10 19:39:18 发布 · 111 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++

该博客主要介绍了一个编程问题，即给定一个正整数序列和一个正整数，判断序列中是否存在一个数，使得它与给定整数之和能被7整除。程序通过遍历序列并检查条件来解决此问题。示例测试用例解释了不同情况下的输出。

Lucky 7

Time Limit: 1000 msMemory Limit: 65536 KB

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 ak from the sequence such that (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 (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), 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≤ai≤100), 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
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".

Author: WENG, Caizhi

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
using namespace std;
int main()
{
	ios::sync_with_stdio(false);
	
	int T,n,x,y;
	cin>>T;
	while(T--)
	{
		cin>>n;
		cin>>x;
		int f=0;
		
		for(int i=1;i<=n;i++)
		{
			cin>>y;
			if((x+y)%7==0&&(f==0))
			{
				f=1;
			}
			
		}
		
		if(f==1)
		{
			cout<<"Yes"<<endl;
		}
		else
		cout<<"No"<<endl;
		
	}
	
	return 0;
}