Problem Description
KazaQ wears socks everyday.

At the beginning, he has n pairs of socks numbered from 1 to n in his closets.

Every morning, he puts on a pair of socks which has the smallest number in the closets.

Every evening, he puts this pair of socks in the basket. If there are n−1 pairs of socks in the basket now, lazy KazaQ has to wash them. These socks will be put in the closets again in tomorrow evening.

KazaQ would like to know which pair of socks he should wear on the k-th day.

Input
The input consists of multiple test cases. (about 2000)

For each case, there is a line contains two numbers n,k (2≤n≤109,1≤k≤1018).

Output
For each test case, output “Case #x: y” in one line (without quotes), where x indicates the case number starting from 1 and y denotes the answer of corresponding case.

Sample Input
3 7
3 6
4 9

Sample Output
Case #1: 3
Case #2: 1
Case #3: 2

题解如下

找规律即可。规律是 1,2,⋯,nn numbers,1,2,⋯,n−1n−1 numbers,1,2,⋯,n−2,nn−1 numbers,1,2,⋯,n−1n−1 numbers,1,2,⋯,n−2,nn−1 numbers,
⋯
⋯ 。

//0MS 1508K
#include <cstdio>  
#include <cstring>  
#include <algorithm> 
#include <cmath>
typedef long long int ll;
using namespace std;
int main()
{
    int cas=1;
    ll n,k,i;
    while(scanf("%lld%lld",&n,&k)!=EOF){
        printf("Case #%d: ",cas);   cas++;
        if(k<=n){
            printf("%lld\n",k);
            continue;
        }
        else{
            ll wei,danjie=(k-n)/(n-1);
            if((k-n)%(n-1)){
                danjie++;
                wei=(k-n)%(n-1);
            }
            else{
                wei=n-1;
            }
            if(danjie%2==0){
                if(wei<n-1){
                    printf("%lld\n",wei);
                }   
                else{
                    printf("%lld\n",n);
                }
            }
            if(danjie%2==1){
                if(wei<n-1){
                    printf("%lld\n",wei);
                }   
                else{
                    printf("%lld\n",n-1);
                }
            }
        }
    }
    return 0;
}