KazaQ's Socks——个人c++解

╰+哭是因爲堅強的太久メ 2022-06-09 11:18 312阅读 0赞

## 问题 K: KazaQ's Socks ##

## Problem Description ##

**KazaQ** wears socks everyday.  
  
At the beginning, he has  n![Image 1][] pairs of socks numbered from  1![Image 1][] to  n![Image 1][] 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![Image 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![Image 1][]\-th day.

## Input ##

The input consists of multiple test cases. (about  2000![Image 1][])  
  
For each case, there is a line contains two numbers  n,k![Image 1][]  (2≤n≤10![Image 1][]9![Image 1][],1≤k≤10![Image 1][]18![Image 1][])![Image 1][].

## Output ##

For each test case, output " **Case \#x![Image 1][]:y![Image 1][]**" in one line (without quotes), where  x![Image 1][] indicates the case number starting from  1![Image 1][] and  y![Image 1][] 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，

n=3时

n k  y

3 1 1

3 2 2

3 3 3

3 4 1

3 5 2

3 6 1

3 7 3

3 8 1

3 9 2

3 10 1

3 11 3

。。。。。。

#include<bits/stdc++.h>//KazaQ's Socks
    using namespace std;
    long long int k;
    int main()
    {
        long long int n,x=1,y;
        while(scanf("%lld %lld",&n,&k)!=EOF)
        {
            if(k<=n)//k<=n时，y=k；
                y=k;
            else
            if((k-n)%(n-1)==0)//判断是否到达了柜子里最后一双袜子
            {
                if((k-n)/(n-1)%2==0)//如果是偶数次到达柜子里最后一双袜子
                    y=n;
                else//如果是奇数次
                    y=n-1;
            }
            else//判断没有到达柜子里最后一双袜子
            y=(k-n)%(n-1);
            printf("Case #%lld: %lld\n",x,y);
            x++;
        }
        return 0;
    }

##     ##

[Image 1]: