【多校联合】（HDU6043）KazaQ's Socks_every evening, he puts this pair of socks in the b-优快云博客

本文针对 HDU6043 KazaQ's Socks 的算法问题进行了详细解析，介绍了如何通过数学推导解决该问题，包括对输入数据进行处理的方法，以及根据特定条件输出结果的实现思路。

【多校联合】（HDU6043）KazaQ’s Socks

一条纯粹的水题，记录下只是因为自己错的太多而已。
原因在于对数据的细节的把握不佳。

原题

KazaQ’s Socks

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 131072/131072 K (Java/Others)

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

3 7
3 6
4 9

Sample Output

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

代码

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

typedef long long ll;


int main()
{
    ll n,k,kase=0;
    while(cin>>n>>k)
    {
        cout<<"Case #"<<++kase<<": ";
        if(k>n)
        {
            k-=n;
            int flag=(k/(n-1)+1)%2;
            if(flag)
                cout<<int(k%(n-1)==0?n:k%(n-1))<<endl;
            else cout<<int(k%(n-1)==0?n-1:k%(n-1))<<endl;
        }
        else cout<<k<<endl;
    }
    return 0;
}