题目描述

You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.

For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1,2,4,5,7,8,10,11,13…. The 7-th number among them is 10.

Input
The first line contains an integer t (1≤t≤1000) — the number of test cases in the input. Next, t test cases are given, one per line.

Each test case is two positive integers n (2≤n≤109) and k (1≤k≤109).

Output
For each test case print the k-th positive integer that is not divisible by n.

样例

6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1

10
15
1999999999
113
1000000001
1

首先我们先看1~k中第k个数是k如果n大于k 直接输出k 如果n小于k有一些数是会被n整除的
然后 我们再看1~n中 有n-1个数不会被n整除 所以 咱们就可以划分出很多 区间元素为n-1个的区间 咱们要求第k个数
让k除以n-1就可以判断k在那个区间 再让k%(n-1)就可以判断k是哪个区间第几个数 然后咱们在让n扩大k/(n-1)倍这里要注意当k%(n-1)等于0时候要特判一下 第k个数正好是第(n-1)个区间最后一个元素

#include<bits/stdc++.h>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        int n,k;
        cin>>n>>k;
        if(n>k)
        {
            cout<<k<<endl;
        }
        else{
            int a=k/(n-1);
            int b=k%(n-1);
            if(b==0)
            {
                cout<<n*a-1<<endl;
            }
            else{
                cout<<n*a+b<<endl;
            }
        }
    }
    return 0;
}