KazaQ's Socks

KazaQ's Socks

KazaQ wears socks everyday.

At the beginning, he has nn pairs of socks numbered from 11 to nn 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−1n−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 kk-th day.

Input

The input consists of multiple test cases. (about 20002000)

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

Output

For each test case, output " Case #xx: yy" in one line (without quotes), where xx indicates the case number starting from 11 and yy 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

官方题解：找规律即可。规律是 $\underbrace{1, 2, \cdots, n}_{n\text{ numbers}},$ $\underbrace{1, 2, \cdots, n - 1}_{n - 1\text{ numbers}},$ $\underbrace{1, 2, \cdots, n - 2, n}_{n - 1\text{ numbers}},$ $\underbrace{1, 2, \cdots, n - 1}_{n - 1\text{ numbers}},$ $\underbrace{1, 2, \cdots, n - 2, n}_{n - 1\text{ numbers}},$ $\cdots$ 。

开始规律找错了（orz),结果队友自己找出来了。

其实自己模拟一下规律还是很明显的。

以3为例；

1 2 3 1 2 1 3 1 2 1 3

以4为例

1 2 3 4 1 2 3 1 2 4 1 2 3 1 2 4

就是n-k之后出现了一个奇偶的循环

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;

typedef long  long ll;

ll n,k;
int main()
{
    int kase=1;
    while(scanf("%lld%lld",&n,&k)!=EOF)
    {
        if(n>=k)
            printf("Case #%d: %d\n",kase++,k);
        else
        {
            ll a=(k-n)/(n-1);
            ll b=(k-n)%(n-1);
            if(a&1)
            {
                if(b==0)
                    printf("Case #%d: %d\n",kase++,n-1);
                else
                    printf("Case #%d: %d\n",kase++,b);
            }
            else
            {
                if(b==0)
                    printf("Case #%d: %d\n",kase++,n);
                else
                    printf("Case #%d: %d\n",kase++,b);
            }
        }
    }
    return 0;
}