hdu-多校训练赛-KazaQ's Socks

KazaQ's Socks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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 73 64 9

 

Sample Output

Case #1: 3Case #2: 1Case #3: 2

题意：

KazaQ 柜子里放着n双袜子，标号分别从1到n，KazaQ 每天早上都会从柜子里拿一双编号最小的袜子穿，夜晚把白天穿的袜子放到一个篮子里，等篮子里的袜子积攒到n-1双后，他就不得不洗，等第二天夜晚把洗完的袜子放到柜子里，给出n和k，求第k天KazaQ 穿的袜子编号是多少。

分析：是一道规律题。如果k小于n，那么第k天的袜子标号一定是k；如果k大于n，从第n天开始，只有编号最大的和次大的有变动

一个简单的规律；；  Case：3    1 2 3 4， 1 2 3， 1 2 4， 1 2 3， 1 2 4......

 

#include<cstdio>using namespace std;int main(){   long long int n,k;    int t=0;    while(~scanf("%lld %lld",&n,&k)){        t++;        printf("Case #%d: ",t);        if(k<n) printf("%lld\n",k);        else{            if((k-n)%(n-1)==0){                if((k-n)/(n-1)%2==0) printf("%lld\n",n);                else                    printf("%lld\n",n-1);            }            else                printf("%d\n",(k-n)%(n-1));        }    }}