| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 5957 | Accepted: 1833 |
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.
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
题目大意就是给出n和k求出第k个与n互素的数
如果知道欧几里德算法的话就应该知道gcd（b×t+a，b）=gcd（a，b） （t为任意整数）
则如果a与b互素，则b×t+a与b也一定互素，如果a与b不互素，则b×t+a与b也一定不互素
故与m互素的数对m取模具有周期性，则根据这个方法我们就可以很快的求出第k个与m互素的数
假设小于m的数且与m互素的数有k个，其中第i个是ai，则第m×k+i与m互素的数是k×m+ai
这道题这样做并不是最优的，网上说可以用欧拉函数+容斥原理+二分枚举可做，我暂时还不知道这样做的思想
如果用我的方法做的话，时间上的花费比较大
#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;
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;
}