HDU 6043 KazaQ's Socks 【规律】

KazaQ's Socks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 462    Accepted Submission(s): 300

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

题目链接：

2017多校训练第一场 1011

题目大意：

KazaQ有n双袜子，这些袜子的编号为1-n，他穿袜子时会从衣橱里选编号最小的那一双来穿。并且每天早上穿，晚上脱下来放到篮子里存着，当篮子里的袜子数量达到n-1双时，他会将它们清洗并在第二天将它们放回衣橱里。现在知道他有n双袜子，编号为1-n，问第k天他穿的袜子编号为多少。

解题思路：

经过观察发现，他穿的袜子编号遵循以下规律：1、2……n、1、2……n-1、1、2……n-2、n、1、2……n-1、1、2……n-2、n……

即前n天中第k天编号就是k，后面每n-1天编号都是1—n-1、1—n-2、n，这样重复。然后根据这个规律来做就好了。

Mycode:

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int MAX = 10005;const int MOD = 1e9+7;const int INF = 0x3f3f3f3f;int n, cas, ans;LL k;int main(){    cas = 0;    while(~scanf("%d%lld",&n,&k))    {        if(n >= k)            ans = k;        else        {            k -= n;            int res = k % (n-1);            if(!res)            {                int tt = k / (n-1);                if(tt & 1)                    ans = n-1;                else                    ans = n;            }            else                ans = res;        }        printf("Case #%d: %d\n",++cas, ans);    }    return 0;}