HDU_6043_KazaQ's Socks

KazaQ's Socks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1890    Accepted Submission(s): 1061

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

 

 

Source

2017 Multi-University Training Contest - Team 1

 

• 循环，找规律
• 前n个顺序不变，从第n+1个开始长度为2n-2的循环
• 拿n=4为例
• 1 2 3 4 /1 2 3 /1 2 4 /1 2 3 /1 2 4 ....

 

 

#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
typedef long long           LL ;
typedef unsigned long long ULL ;
const int    maxn = 1e5 + 10   ;
const int    inf  = 0x3f3f3f3f ;
const int    npos = -1         ;
const int    mod  = 1e9 + 7    ;
const int    mxx  = 100 + 5    ;
const double eps  = 1e-6       ;
const double PI   = acos(-1.0) ;

int main(){
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    LL T=0, n, k, ans;
    while(~scanf("%lld %lld",&n,&k)){
        if(k<=n){
            ans=k;
        }else{
            k-=n;
            LL m=k/(n-1);
            if(m*(n-1)<k)m++;
            k=k-(n-1)*(m-1);
            ans=k;
            if(!(m&1))
                if(k==n-1)
                    ans++;
            
        }
        printf("Case #%lld: %lld\n",++T,ans);
    }
    return 0;
}