hdu6043KazaQ's Socks（高校1）

KazaQ’s Socks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1149 Accepted Submission(s): 684

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

例子 3 7
则会出现:1 2 3 1 2 1 3 1 2 1 3 1 2 1 3 1 2 1 3……………….

例子：4 6
出现:1 2 3 4 1 2 3 1 2 4 1 2 3 1 2 4 1 2 3 1 2 4.。。。。。

/*所以得出规律，当k<n时，直接输出k，当k>=n时，k去掉前n个，也就是第一次到循环点的袜子数，然后剩下的数量取余n-1，如果得到0，且除以n-1刚好为奇数那么证明k刚好在剩下的袜子循环点上，则直接输出n-1，刚好取余且为偶数则输出n,其余输出除出来的值取余n-1*/#include <iostream>#define LL long longusing namespace std;int main(){    LL n,k;    LL ca=1;    while(cin>>n>>k)    {        cout<<"Case #"<<ca++<<": ";        if(k<=n)            cout<<k<<endl;        else        {            k-=n;            LL t=2*(n-1);            LL tem=n-1;            LL m=k%t;            if(m==0)                cout<<n<<endl;            else if(m==tem)            {                cout<<n-1<<endl;            }            else            {                cout<<m%tem<<endl;            }        }    }    //cout<<0<<endl;    return 0;}