2017 Multi-University Training Contest

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. (about2000)

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： 给定一个n,k (2≤n≤109,1≤k≤1018)      一个人有n双袜子， 编号1-n， 每天早上穿衣柜里编号最下的一双， 晚上放到洗衣篮里， 当衣柜里只有
一双袜子的时就将之前的袜子洗掉， 并在第二天晚上放回衣柜里， 问第k天穿的是第几双袜子。
思路， 找规律的签到题， 举个例子 n=5时 穿袜子的规律就是 1 2 3 4 5   1 2 3 4    1 2 3 5  1 2 3 4   1 2 3 5
可以看出， 前n天不用管， 之后每n-1天就是循环节， 每一个循环节 前n-2天不变， 之后第n-1天跟循环的奇偶有关
#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<algorithm>#include<queue>#include<map>using namespace std;typedef long long LL;LL n, k;int main(){    int Case = 0;    while(~scanf("%lld %lld", &n, &k)){        printf("Case #%d: ", ++Case);        if(k <= n) printf("%lld\n", k);        else{            k -= n;            LL cnt = k / (n - 1);            if(k % (n - 1)) cnt++;            k -= (n - 1) * (cnt - 1);            if(k != n - 1) printf("%lld\n", k);            else {                if(cnt & 1) printf("%lld\n", n - 1);                else printf("%lld\n", n);            }        }    }    return 0;}