HDU6043-KazaQ's Socks

最新推荐文章于 2017-09-14 19:50:09 发布

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探讨了一个人每天从箱子中取出最小编号袜子并晚上放入篮子的问题。当篮子中有n-1双袜子时，此人会将袜子清洗后再次放入箱子。问题是求第k天他会穿哪双袜子。

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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

Sample Output

  
   Case #1: 3
Case #2: 1
Case #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 long
const 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;
}