题目传送：http://acm.hdu.edu.cn/showproblem.php?pid=6043
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
题意：这个人每天早上会从柜子拿一双序号最小的袜子，晚上扔进脏衣篓。 他会等到干净的袜子只剩一双的时候才去洗脏衣篓的袜子，给你袜子的总数n和第几天k，问你第k天穿的袜子序号为几。
题解：可以通过分析n=3和n=4的情况来推断，打括号的说明是一起洗的。
n=3：(1 2)( 3 1 )(2 1)( 3 1)( 2 1)( 3 1).....
n=4: (1 2 3) (4 1 2) (3 1 2) (4 1 2) (3 1 2) (4 1 2) (3 1 2)......
除去第一个括号，每一个括号的开头都是仅剩的最后一双袜子，然后会从上一括号中洗好的再选（0~n-2）双，所以就有了这样的规律。
#include<iostream>
#include<cstdio> using namespace std; int main()
{
int t=;
long long n,k,res;
while(~scanf("%lld%lld",&n,&k))
{
if(k<=n)
{
res=k;
}
else
{
k-=n;
int q=k%(n-);
if(q==)
if(k/(n-)%==)
res=n;
else
res=n-;
else
res=q;
}
printf("Case #%d: %lld\n",t++,res);
}
return ;
}