17多校contest 1 - 1011 KazaQ's Socks （规律

最新推荐文章于 2017-10-04 18:44:00 发布

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本文介绍了一个关于KazaQ日常穿袜子习惯的问题，并通过寻找规律解决了一个数学问题。该问题涉及如何确定在第k天KazaQ会穿哪一双袜子。文中提供了一段AC代码，用于解决此问题。

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KazaQ’s Socks

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 7
3 6
4 9

Sample Output

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

Hint

题意

KazaQ每天穿袜子一开始，他的衣柜里有n对袜子从1到n。每天早晨，他穿上一双袜子，衣柜里数量最少。每天晚上，他把这双袜子放在篮子里。如果篮子里有n-1对袜子，懒惰的KazaQ就要洗了。这些袜子将在明天晚上再次放在衣橱里。 KazaQ想知道他在第k天应该穿什么袜子

题解：

找两组规律就出来的多校签到题2/1

AC代码

#include <bits/stdc++.h>
using namespace std;

#define LL long long

int main()
{
    int res = 0;
    LL n, k;
    while(~scanf("%lld%lld",&n,&k)) {
        printf("Case #%d: ",++res);
        if(k <= n) printf("%lld\n",k);
        else {
            k -= n;
            k %= (2*(n-1));
            if(k == 0) {
                printf("%lld\n",n);
            }
            else {
                if(k>=1 && k<=n-1)
                    printf("%lld\n",k);
                else
                    printf("%lld\n",k+1-n);
            }
        }
    }
return 0;
}