C. Raspberries

time limit per test

2 seconds

memory limit per test

256 megabytes

You are given an array of integers a1,a2,...,ana1,a2,...,an and a number kk (2≤k≤52≤k≤5). In one operation, you can do the following:

  • Choose an index 1≤i≤n1≤i≤n,
  • Set ai=ai+1ai=ai+1.

Find the minimum number of operations needed to make the product of all the numbers in the array a1⋅a2⋅...⋅ana1⋅a2⋅...⋅an divisible by kk.

Input

Each test consists of multiple test cases. The first line contains a single integer tt (1≤t≤1041≤t≤104) --- the number of test cases. Then follows the description of the test cases.

The first line of each test case contains two integers nn and kk (2≤n≤1052≤n≤105, 2≤k≤52≤k≤5) --- the size of the array aa and the number kk.

The second line of each test case contains nn integers a1,a2,...,ana1,a2,...,an (1≤ai≤101≤ai≤10).

It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105.

Output

For each test case, output the minimum number of operations needed to make the product of all the numbers in the array divisible by kk.

Example

Input

Copy

复制代码

15

2 5

7 3

3 3

7 4 1

5 2

9 7 7 3 9

5 5

5 4 1 2 3

7 4

9 5 1 5 9 5 1

3 4

6 3 6

3 4

6 1 5

3 4

1 5 9

4 4

1 4 1 1

3 4

3 5 3

4 5

8 9 9 3

2 5

1 6

2 5

10 10

4 5

1 6 1 1

2 5

7 7

Output

Copy

复制代码
2
2
1
0
2
0
1
2
0
1
1
4
0
4
3

Note

In the first test case, we need to choose the index i=2i=2 twice. After that, the array will be a=[7,5]a=[7,5]. The product of all the numbers in the array is 3535.

In the fourth test case, the product of the numbers in the array is 120120, which is already divisible by 55, so no operations are needed.

In the eighth test case, we can perform two operations by choosing i=2i=2 and i=3i=3 in any order. After that, the array will be a=[1,6,10]a=[1,6,10]. The product of the numbers in the array is 6060.

解题说明:此题是一道数学题,由于K范围很小,可以分类讨论。当k=4的时候存在给两个数都加1的情况,其他情况下的k都只可以给一个数一直加1直到这个数可以整除k。

cpp 复制代码
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int INF = 0x3f3f3f3f;
int main()
{
	int T; 
	cin >> T;
	while (T--)
	{
		int n, k; 
		cin >> n >> k;
		int a[100004];
		int ans = INF;
		int cal = 1;
		for (int i = 1; i <= n; i++)
		{
			cin >> a[i];
			cal *= a[i];
			if (a[i] % k == 0) 
			{
				ans = 0;
			}
			int t = a[i] / k + 1;
			ans = min(ans, t * k - a[i]);
		}
		if (k == 4)
		{
			int cnt1 = 0, cnt2 = 0;
			for (int i = 1; i <= n; i++)
			{
				switch (a[i] % 4)
				{
				case 1:
					cnt1++;
					break;
				case 2:
					cnt2++;
					break;
				case 3:
					break;
				default:
					ans = 0;
					break;
				}
			}
			if (cnt1 >= 2)
			{
				ans = min((int)2, ans);
			}
			if (cnt2 >= 2)
			{
				ans = 0;
			}
			if (cnt1 >= 1 && cnt2 >= 1)
			{
				ans = min((int)1, ans);
			}
		}
		cout << ans << endl;
	}
	return 0;
}
相关推荐
初见无风19 分钟前
2.5 Lua代码中string类型常用API
开发语言·lua·lua5.4
做运维的阿瑞28 分钟前
用 Python 构建稳健的数据分析流水线
开发语言·python·数据分析
左师佑图32 分钟前
综合案例:Python 数据处理——从Excel文件到数据分析
开发语言·python·数据分析·excel·pandas
陌路201 小时前
C23构造函数与析构函数
开发语言·c++
_OP_CHEN2 小时前
C++进阶:(二)多态的深度解析
开发语言·c++·多态·抽象类·虚函数·多态的底层原理·多态面试题
CsharpDev-奶豆哥2 小时前
JavaScript性能优化实战大纲
开发语言·javascript·性能优化
小妖同学学AI2 小时前
Rust 深度解析:变量、可变性与所有权的“安全边界”
开发语言·安全·rust
2301_764441332 小时前
基于python构建的低温胁迫实验
开发语言·python
ICT系统集成阿祥2 小时前
华为CloudEngine系列交换机堆叠如何配置,附视频
开发语言·华为·php
wjs20242 小时前
C++ 基本语法
开发语言