题目描述：
Suppose you are living with two cats: A and B. There are n napping spots where both cats usually sleep.
Your cats like to sleep and also like all these spots, so they change napping spot each hour cyclically:
Cat A changes its napping place in order: n,n−1,n−2,…,3,2,1,n,n−1,… In other words, at the first hour it’s on the spot n and then goes in decreasing order cyclically;
Cat B changes its napping place in order: 1,2,3,…,n−1,n,1,2,… In other words, at the first hour it’s on the spot 1 and then goes in increasing order cyclically.
The cat B is much younger, so they have a strict hierarchy: A and B don’t lie together. In other words, if both cats’d like to go in spot x then the A takes this place and B moves to the next place in its order (if x<n then to x+1, but if x=n then to 1). Cat B follows his order, so it won’t return to the skipped spot x after A frees it, but will move to the spot x+2 and so on.
Calculate, where cat B will be at hour k?
Input
The first line contains a single integer t (1≤t≤104) — the number of test cases.
The first and only line of each test case contains two integers n and k (2≤n≤109; 1≤k≤109) — the number of spots and hour k.
Output
For each test case, print one integer — the index of the spot where cat B will sleep at hour k.
You live a happy life with an orange cat and a Persian cat. There are n places in your home where cats can sleep.
Your cat likes to sleep in any one place, so they will periodically change places every hour:
· The orange cat changes its sleeping position in the order of n, n-1, n-2, … , 3, 2, 1, n, n-1, …;
· Persian cats change their sleeping positions in the order of 1, 2, 3, … , n-1, n, 1, 2, ….
The orange cat is much larger than the Persian cat and looks very scary, so when the orange cat and the Persian cat happen to want to sleep in the same place, the Persian cat moves out of the way and moves to the next position in his order.
Where is your Persian cat asleep after k hours?
There is a t in the first line of Input , indicating that there are t groups of inputs. 1 <= t <= 10000
Next there are t lines, each line has two numbers, n and k. 2 <= n <= 1000000000; 1 <= k <= 1000000000
Output
outputs t lines, each with an integer representing the location of your Persian cat.
Example
Input
7
2 1
2 2
3 1
3 2
3 3
5 5
69 1337
Output
1
2
1
3
2
2
65
Note
In the first and second test cases n=2, so:
at the 1-st hour, A is on spot 2 and B is on 1;
at the 2-nd hour, A moves to spot 1 and B — to 2.
If n=3 then:
at the 1-st hour, A is on spot 3 and B is on 1;
at the 2-nd hour, A moves to spot 2; B’d like to move from 1 to 2, but this spot is occupied, so it moves to 3;
at the 3-rd hour, A moves to spot 1; B also would like to move from 3 to 1, but this spot is occupied, so it moves to 2.
In the sixth test case:
A’s spots at each hour are [5,4,3,2,1];
B’s spots at each hour are [1,2,4,5,2].
Problem-solving ideas:
First of all, we know that the position of the orange cat will not change due to any situation, but the Persian cat will change because of the orange cat, and when the number of positions is even, the Persian cat will not change, only the number of positions n It will only change when it is an odd number, and by looking up the law, we found that we can use n/2 to find out how many times the position will be met once, so we can find the total number of steps taken by the Persian cat, and then use the modulo The way to find out the final position of the Persian cat, without further ado, go directly to the code:
#include <iostream>
#include <cstdio>
using namespace std;
int n,t,k;
int main () {
cin >> t;
while (t--) {
cin >> n >> k;
if(n%2 == 0)//偶数的情况
printf("%d\n",k%n ? k%n : n);//如果取模为0则波斯猫的位置为n,否则为n%k
else {
int a = n/2;//求出除了第一个位置外每次经过多少个位置会相遇
int m = (k-1)/a;//找到相遇的次数，即波斯猫少走的步数
int p = (m+k) % n;//用波斯猫的总步数来取模找到波斯猫的位置
if(p == 0)
cout << n << "\n";
else
cout << p << "\n";
}
}
return 0;
}