Title:
solution:
The problem is to find the position of the kth second, that is, the position after k − 1 steps. When n is an even number, two cats will not meet. When n is an odd number: meet once every n 2 seconds, each time b cat The position of is increased by 1, then cat b increases by n 2 + 1 from the beginning to the meeting position. Set x = n / 2, then ans = (kx ∗ (x + 1) + k% x)% n + 1 The position of k seconds, that is, the position after k-1 steps. \\ When n is an even number, two cats will not meet. \\ When n is an odd number: \\ every \frac(n)(2) seconds Encounter once, each time you encounter, the position of cat b increases by 1, \\ Then cat b increases from the beginning to the position of the encounter \frac(n)(2)+1, \\ Set x=n/2, then ans=( \frac{k}{x}*(x+1)+k\%x)\%n+1 Topic entry request of k sec the position set ,That go k−A step of after the bit is set .When n is an even number of the time ,Liang Zhi cats do not will phase encounter .When n is an odd number when:each2nSecond phase in case of a second ,Each sub- phase encounter ,b cat 's position set by adding 1 ,That it b cat from open start to phase in case of position set by adding a2n+1,Let x=n/2,Then a n s=(xk∗(x+1)+k%x)%n+1
code：
#include <bits/stdc++.h>
#define int long long
using namespace std;
int n,k;
void solve(){
cin>>n>>k;
k--;
if(n%2==0){
cout<<k%n+1<<endl;
}else{
int x=n/2;
int ans=(k/x*(x+1)+k%x)%n+1;
cout<<ans<<endl;
}
}
signed main(){
ios::sync_with_stdio(0);
int T;cin>>T;
while(T--){
solve();
}
return 0;
}