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
侃侃_天下2 天前
最终的信号类
开发语言·c++·算法
echoarts2 天前
Rayon Rust中的数据并行库入门教程
开发语言·其他·算法·rust
Aomnitrix2 天前
知识管理新范式——cpolar+Wiki.js打造企业级分布式知识库
开发语言·javascript·分布式
每天回答3个问题2 天前
UE5C++编译遇到MSB3073
开发语言·c++·ue5
伍哥的传说2 天前
Vite Plugin PWA – 零配置构建现代渐进式Web应用
开发语言·前端·javascript·web app·pwa·service worker·workbox
小莞尔2 天前
【51单片机】【protues仿真】基于51单片机的篮球计时计分器系统
c语言·stm32·单片机·嵌入式硬件·51单片机
小莞尔2 天前
【51单片机】【protues仿真】 基于51单片机八路抢答器系统
c语言·开发语言·单片机·嵌入式硬件·51单片机
liujing102329292 天前
Day03_刷题niuke20250915
c语言
我是菜鸟0713号2 天前
Qt 中 OPC UA 通讯实战
开发语言·qt
JCBP_2 天前
QT(4)
开发语言·汇编·c++·qt·算法