杭电 KazaQ's Socks

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 thek-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 73 64 9

 

Sample Output

Case #1: 3Case #2: 1
Case #3: 2
题意：KazaQ 有n双袜子，按1到n编号放在柜子里，他早上选一双穿，号码小的优先，晚上把穿的袜子放篮子里，当篮子里的袜子达到n-1双时，他会洗袜子，第二天晚上会把洗过的袜子放到柜子里，问第k天他穿的是哪双袜子？
n=3时
n  k  ans
3  1  1
3  2  2
3  3  3
3  4  1
3  5  2
3  6  1
3  7  3
3  8  1
3  9  2
3  10  1
3  11  3
。。。。。。
n=4时
n  k  ans
4  1  1
4  2  2
4  3  3
3  4  4
4  5  1
4  6  2
4  7  3
4  8  1
4  9  2
4  10  4
4  11  1
4  12  2
4  13  3
4  14  1
4  15  2
4  16  4
。。。。。。
观察规律：前n天，第i天袜子是i,之后,天数-n,即i-n, n-1为周期，第i天， (i-n)%(n-1) 不为0， 袜子为(i-n)%(n-1)，否则  (i-n)/(n-1)为奇数  袜子n-1 ；
(i-n)/(n-1)为偶数，袜子为n   
ans=
思路：
AC代码如下：
#include <iostream>using namespace std;typedef long long int ll;int main(){    ll n,k;    int ji=1;    while(cin>>n>>k)    {        cout<<"Case #"<<ji<<": ";        ji++;        if(k<=n)        {            cout<<k<<endl;            continue;        }        else{            ll m=(k-n)%(n-1);            if(m==0) m=n;            ll rare=(k-n)/(n-1);            if(m<n-1)            {                cout<<m<<endl;            }            else            {                if(rare%2) cout<<n-1<<endl;                else cout<<n<<endl;            }        }    }    return 0;}