2017杭电多校联赛 1011 KazaQ's Socks（找规律）HDU 6043

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题目大意：有N双袜子，每天穿一双，穿到n天时把前n-1 双袜子都洗了，然后又按顺序穿，问第 k 天是穿的哪双袜子解题思路：其实就是一个规律比如四双袜子，他穿的顺序就是1234123124123124.。。AC代码

#include <iostream>
#include <cstdio>
typedef long long ll;
using namespace std;
ll n,k;
int main()
{
int ans=1;
while(~scanf("%lld%lld",&n,&k))
{
printf("Case #%d: ",ans);
ans++;
if(k<=n)
printf("%lld\n",k);
else
{
k=k-n;
ll p=k%(n*2-2);
if(p==n-1)
printf("%lld\n",n-1);
else if(p==0)
printf("%lld\n",n);
else
{
if(p>=1&&p<n-1)
printf("%lld\n",p);
else
printf("%lld\n",p-n+1);
}
}
}
return 0;
}