题目来源

一、题目

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?

两只小猫A和B，有n个窝，A猫按n,n-1,…2,1,n,n-1,n-2…的顺序挪窝，B猫按1,2…n-1,n,1,2,…,n的顺序挪窝，每小时挪一次。假如两猫相遇，B猫会挪到下一个位置。问:在第k小时B猫在哪？

二、输入

The first line contains a single integer t (1≤t≤1e4) — the number of test cases.
The first and only line of each test case contains two integers n and k (2≤n≤1e9; 1≤k≤1e9) — the number of spots and hour k.

第一行测试样例数t
第二行窝数n和小时k

三、输出

For each test case, print one integer — the index of the spot where cat B will sleep at hour k.

输出B在第k小时所在的窝的编号

四、样例

input:

output:

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].

五、思路 + 代码

一道简单题，很容易想到用模拟的方法去解，但是1e9的范围用o(n)复杂度的方法的话极大可能就超时了，还是要找数学规律。
n分两种情况，奇数和偶数。偶数时，两只猫不会相遇；奇数时，每走n/2步相遇一次。分类讨论即可。

代码如下：

#include<iostream>
using namespace std;
int n, k, ans;
void solve() {cin >> n >> k;if (n & 1) {int meet = (k - 1) / (n >> 1);int step = k - 1;ans = 1 + (step + meet) % n;}else {ans = k % n == 0 ? n : k % n;}cout << ans << endl;
}
int main() {int t;cin >> t;while (t--) {solve();}return 0;
}

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B. Cat Cycle

一、题目

二、输入

三、输出

四、样例

五、思路 + 代码

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