多校1 K KazaQ's Socks

最新推荐文章于 2019-03-11 21:49:00 发布

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本文介绍了一个关于KazaQ每天选择袜子的问题，通过特定规律计算第k天穿哪双袜子。输入包括多组测试案例，每组包含整数n和k，输出则是第k天穿的袜子编号。

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KazaQ's Socks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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

这道题我没看，主要是另外两个队友打的，他们很快就知道规律了，我想应该不会很难，代码是斌斌打的~~~

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
int a[11111];
int main()
{
    int cas=0;
    long long int n,k;
    long long int cmp,res;
    while(cin>>n>>k)
    {
        cas++;
        cmp=2*(n-1);
        res=(k-n)%cmp;
        if(n>=k)
            printf("Case #%d: %d\n",cas,k);
        else if(res==0)
            printf("Case #%d: %d\n",cas,n);
        else if(res<=cmp/2)
            printf("Case #%d: %d\n",cas,res);
        else
            printf("Case #%d: %d\n",cas,res-cmp/2);
    }
    return 0;
}