正文
【多校联合】（HDU6043）KazaQ's Socks
小程序：扫一扫查出行【扫一扫了解最新限行尾号】
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【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
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.
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).
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.
3 7
3 6
4 9
Case #1: 3
Case #2: 1
Case #3: 2
#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;
}
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