Happy 2006
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 6940 | Accepted: 2192 |
Description
Two positive integers are said to be relatively prime to each other if the Great Common Divisor (GCD) is 1. For instance, 1, 3, 5, 7, 9...are all relatively prime to 2006.
Now your job is easy: for the given integer m, find the K-th element which is relatively prime to m when these elements are sorted in ascending order.
Now your job is easy: for the given integer m, find the K-th element which is relatively prime to m when these elements are sorted in ascending order.
Input
The input contains multiple test cases. For each test case, it contains two integers m (1 <= m <= 1000000), K (1 <= K <= 100000000).
Output
Output the K-th element in a single line.
Sample Input
2006 1 2006 2 2006 3
Sample Output
1 3 5 #include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int pri[1000000];
int gcd ( int a , int b )
{
return b == 0 ? a : gcd ( b , a % b ) ;
}
int main()
{
int m , k ;
while ( cin >> m >> k )
{
int i , j ;
for ( i = 1 , j = 0 ; i <= m ; i ++ )
if ( gcd ( m , i ) == 1 )
pri [ j ++ ] = i ;
if ( k%j != 0)
cout <<k/j * m +pri[k%j-1] << endl; //注意i与m互素，则i+km与m也互素，展转相除法可得出
else//要特别考虑k%j=0的情况，因为数组是从0开始的，第i个对应的是pri[i-1]
cout << (k/j-1)*m+pri[j-1] << endl ;
}
return 0;
}