Live Love

https://pintia.cn/problem-sets/1036903825309761536/problems/1041155943483625472

A Live Love

DreamGrid is playing the music game Live Love. He has just finished a song consisting of n notes and got a result sequence A​1​​,A​2​​,...,A​n​​ (A​i​​∈ {PERFECT, NON-PERFECT}). The score of the song is equal to the max-combo of the result sequence, which is defined as the maximum number of continuous PERFECTs in the sequence.

Formally speaking, max-combo(A)=max { k | k is an integer and there exists an integer i (1≤i≤n−k+1) such that A​i​​=A​i+1​​=A​i+2​​=...=A​i+k−1​​= PERFECT }. For completeness, we define max(∅)=0.

As DreamGrid is forgetful, he forgets the result sequence immediately after finishing the song. All he knows is the sequence length n and the total number of PERFECTs in the sequence, indicated by m. Any possible score s he may get must satisfy that there exists a sequence A​′​​ of length n containing exactly mPERFECTs and (n−m) NON-PERFECTs and max-combo(A​′​​)=s. Now he needs your help to find the maximum and minimum s among all possible scores.

Input

There are multiple test cases. The first line of the input contains an integer T (1≤T≤100), indicating the number of test cases. For each test case:

The only line contains two integers n and m (1≤n≤10​3​​, 0≤m≤10​3​​, m≤n), indicating the sequence length and the number of PERFECTs DreamGrid gets.

Output

For each test case output one line containing two integers s​max​​ and s​min​​, indicating the maximum and minimum possible score.

Sample Input

5
5 4
100 50
252 52
3 0
10 10

Sample Output

4 2
50 1
52 1
0 0
10 10

Hint

Let's indicate a PERFECT as P and a NON-PERFECT as N.

For the first sample test case, the sequence (P,P,P,P,N) leads to the maximum score and the sequence (P,P,N,P,P) leads to the minimum score.

 

题意是给出所有的长度和perfect的个数，求可能的最大分数和最小分数，最大分数肯定是所有的perfect都连在一起的情况，最小分数是刚好所有的perfect被平均分开的情况。

#include<stdio.h>
int main()
{
	int t,n,m,max,min;
	scanf("%d",&t);;
	while(t--)
	{
		scanf("%d%d",&n,&m);
		max=m;
		m=n-m+1;
		min=n/m;
		printf("%d %d\n",max,min);
	}
	return 0;
}