HDU 6043 KazaQ's Socks（暑期训练1011）

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探讨了KazaQ日常更换及清洗袜子的规律，通过数学建模找到第k天所穿袜子编号的算法解决方案。

KazaQ's Socks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 448 Accepted Submission(s): 290

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 #

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 7
3 6
4 9

Sample Output

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

找规律。每到n-1双袜子就要洗一下，所以n-1是一个循环。因为每次洗袜子时穿的必是第n或n-1双，所以n与n-1形成循环，所以循环为2*（n-1）。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
long long n,k;
int main()
{
    int cnt=0;
    while(~scanf("%I64d %I64d",&n,&k))
    {
        cnt++;
        if(k<=n)
        {
            printf("Case #%d: %I64d\n",cnt,k);
            continue;
        }
        long long a=(n-1)*2,b=k-n;
        long long m=b%a;
        //printf("m=%I64d\n",m);
        if(m<n-1&&m>0)
            printf("Case #%d: %I64d\n",cnt,m);
        else if(m==n-1)
            printf("Case #%d: %I64d\n",cnt,n-1);
        else if(m>n-1&&m<2*n-2)
            printf("Case #%d: %I64d\n",cnt,m-(n-1));
        else if(m==0)
            printf("Case #%d: %I64d\n",cnt,n);
    }
    return 0;
}