poj 2773 Happy 2006（求第k个与n互质的数）

Happy 2006

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8779		Accepted: 2913

DescriptionTwo 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.InputThe input contains multiple test cases. For each test case, it contains two integers m (1 <= m <= 1000000), K (1 <= K <= 100000000).OutputOutput the K-th element in a single line.Sample Input

2006 1
2006 2
2006 3

Sample Output

1
3
5

题意：给出m、k，求k个与m互质的数。

思路：我们知道gcd（b×t+a，b）=gcd（a，b）  （t为任意整数），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。

AC代码：

#include <iostream>#include <cmath>#include <cstdlib>#include <cstring>#include <cstdio>#include <queue>#include <ctime>#define ll __int64using namespace std;const int maxn = 1000000;int prime[maxn + 5];bool vis[maxn + 5], flag[maxn + 5];int cnt;void getprime(){memset(vis, false, sizeof(vis));cnt = 0;for(int i = 2; i <= maxn; i++)if(!vis[i]){prime[cnt++] = i;for(int j = 1; j * i<= maxn; j++)vis[j * i] = true;}}int euler(int n){memset(flag, false, sizeof(flag));int ret = n, m = n;for(int i = 0; prime[i] * prime[i] <= n; i++)if(n % prime[i] == 0){ret = ret / prime[i] * (prime[i] - 1);while(n % prime[i] == 0) n /= prime[i];for(int j = prime[i]; j <= m; j += prime[i])flag[j] = true;}if(n > 1){ret = ret / n * (n - 1);for(int j = n; j <= m; j += n)flag[j] = true;}return ret;}int find(int n, int id){int cnt = 0;for(int i = 1; i <= n; i++){if(!flag[i]) cnt++;if(cnt == id) return i;}return -1;}int main(){int m, k;getprime();while(cin>>m>>k){if(m == 1){cout<<k<<endl;continue;}int len = euler(m);cout<<((k - 1)/ len) * m + find(m, (k - 1) % len + 1)<<endl;}return 0;}