KazaQ’s Socks
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 131072/131072 K (Java/Others)
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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.
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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).
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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.
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Sample Input
3 7
3 6
4 9Sample Output
Case #1: 3
Case #2: 1
Case #3: 2?
题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=6043
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分析
题意：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
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
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代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{int res=1,n;ll k;while(~scanf("%d%I64d",&n,&k)){printf("Case #%d: ",res++);if(k<=n)printf("%d\n",k);else{k-=n;int s=(k/(n-1))%2,t=(k-1)%(n-1)+1;if(s)printf("%d\n",t);else printf("%d\n",t==n-1?n:t);}}return 0;
}