HDU6043-KazaQ's Socks

KazaQ's Socks

                                                                       Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                                                 Total Submission(s): 518    Accepted Submission(s): 339

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

 

Source

2017 Multi-University Training Contest - Team 1

 

题意：有个人有n双袜子，一开始袜子袜子全放箱子里，每天早上他会拿出编号最小的袜子来穿，晚上他会将袜子放入篮子，若篮子里的袜子等于你n-1双，则他会将袜子拿去洗，第二天晚上会将洗好的袜子放回箱子里，问第k天他穿哪双袜子

解题思路：找规律后发现，n双袜子的循环节为：除了前n天是1~n，后面的的都是1~n，1~n-1，1~n-2，n

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int main(){    LL n,k;    int cas=0;    while(~scanf("%lld%lld",&n,&k))    {        printf("Case #%d: ",++cas);        if(k<=n) printf("%lld\n",k);        else        {            k-=n;            LL m=2*(n-1);            k%=m;            if(k<=n-1&&k) printf("%lld\n",k);            else if(!k) printf("%lld\n",n);            else printf("%lld\n",k-(n-1));        }    }    return 0;}