HDU 6043 KazaQ's Socks

https://vjudge.net/problem/HDU-6043

 

KazaQ wears socks everyday. 

At the beginning, he has nn pairs of socks numbered from 11 to nn 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−1n−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 kk-th day.

InputThe input consists of multiple test cases. (about 20002000) 

For each case, there is a line contains two numbers n,kn,k (2≤n≤109,1≤k≤1018)(2≤n≤109,1≤k≤1018).OutputFor each test case, output " Case #xx: yy" in one line (without quotes), where xxindicates the case number starting from 11 and yy 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

分析找规律
会有状态循环：
从第一次取完后开始，每两次取完所有之后会开始重复


代码

#include <bits/stdc++.h>
#include <iostream>
#include <cstring>
#include <stack>
#include <cstdlib>
#include <queue>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <string>
#include <vector>
#include <list>
#include <iterator>
#include <set>
#include <map>
#include <utility>
#include <iomanip>
#include <ctime>
#include <sstream>
#include <bitset>
#include <deque>
#include <limits>
#include <numeric>
#include <functional>

#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
#define mod 1000000007
#define sort(a,n,int) sort(a,a+n,less<int>())
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)
using namespace std;

typedef long long ll;
typedef char ch;
typedef double db;

const int maxn=1e5+10;
//ll a[maxn];
int aa[maxn];

int main()
{
    ll m ,n;
    while(cin >> m >> n)
    if(n<=m)cout<<"Case #"<<t<<": "<<n<<endl;
    else
    {
        n -= m;
        n %= (2*m-2);
        if(!n)cout<<"Case #"<<t<<": "<<m<<endl;
        else
        {
            if(n<=m-1) cout<<"Case #"<<t<<": "<<n<<endl;
            else cout<<"Case #"<<t<<": "<<n-m+1<<endl;
        }
    }
    return 0;
}