Educational Codeforces Round 103 (Rated for Div. 2) A. K-divisible Sum

Educational Codeforces Round 103 (Rated for Div. 2) A. K-divisible Sum)

目录

• 题面
• 题目分析
  • AC代码

题面

You are given two integers \(n\) and \(k\).

You should create an array of \(n\) positive integers \(a1,a2,…,an\) such that the sum \((a1+a2+⋯+an)\) is divisible by \(k\) and maximum element in aa is minimum possible.

What is the minimum possible maximum element in \(a\)?

Input

The first line contains a single integer \(t\) \((1≤t≤1000\)) — the number of test cases.

The first and only line of each test case contains two integers \(n\) and \(k\) \((1≤n≤109\))\(1≤; (1≤k≤109)\).

Output

For each test case, print one integer — the minimum possible maximum element in array aa such that the sum \((a1+⋯+an)\) is divisible by \(k\).

Example

input

4
1 5
4 3
8 8
8 17

output

5
2
1
3

Note

In the first test case \(n=1\), so the array consists of one element \(a1\) and if we make \(a1=5\) it will be divisible by \(k=5\) and the minimum possible.

In the second test case, we can create array \(a=[1,2,1,2]\). The sum is divisible by \(k=3\) and the maximum is equal to \(2\).

In the third test case, we can create array \(a=[1,1,1,1,1,1,1,1]\). The sum is divisible by \(k=8\) and the maximum is equal to \(1\).

题目分析

为了使构成的数组的最大值最小，只需要让数组中的元素尽可能相同即让数字都等于n / k，当n / k无法整除时，那么最大值就是 n / k向上取整(ceil)。但是，当\(n>k\)时便无法满足，这个时候我们需要让k倍增，使k刚好大于等于n，我们只需要k = k * ceil(k / n)就可以实现。最终输出答案即可。

AC代码

#include<bits/stdc++.h>
using namespace std;
#define io ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
int t;
long long n, k;

int main(){
    io;
    cin >> t;
    while(t--){
	    cin >> n >> k;
		int t = k;
		if(n > k){
			k = k * (ceil((double)n / (double)k));
		}
		long long ans = ceil((double) k / (double)n);
		cout << ans << endl;
    }

    return 0;
}