HDU 6043（2017 多校训练赛1 1011） KazaQ's Socks

2017 Multi-University Training Contest - Team 1 1011

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 #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

 

题意：有n双袜子序号分别为1~n，每天早上穿序号最小的那双，每天晚上放到篮子里，当篮子里袜子数等于n-1时第二天会洗问第k天穿的哪双袜子。

分析：这个其实是找规律，会发现每次变化的只是 n和 n-1
   比如第一个样例 3 7
   那么7天内分别穿的是   1  2   3   1   2   1   3  
   再往后写几天会发现 循环节为   3   1   2   1

AC代码：

#include<stdio.h>
#include<string.h>
int main()
{
long long n,k,cas=0;
while(scanf("%lld%lld",&n,&k)==2)
{
printf("Case #%d: ",++cas);
if(k<=n)
printf("%lld\n",k);
else
{
long long a=(k-n)/(n-1);
long long b=(k-n)%(n-1);
if(a%2)
{
if(b==0)
printf("%lld\n",n-1);
else
printf("%lld\n",b);
}
else
{
if(b==0)
printf("%lld\n",n);
else
printf("%lld\n",b);
}
}
}
}