/*Description
Given n, a positive integer, how many positive integers less than n are relatively prime to n? Two integers a and b are relatively prime if there are no integers x > 1, y > 0, z > 0 such that a = xy and b = xz. Input
There are several test cases. For each test case, standard input contains a line with n <= 1,000,000,000. A line containing 0 follows the last case. Output
For each test case there should be single line of output answering the question posed above. Sample Input
7 12 0
Sample Output
6 4
#include <iostream> #include<cmath> #include <algorithm> using namespace std; int Euler(int n) { int ret=n,p; for( p=2;p*p<=n;p++) { if(n%p==0) { ret=ret/p*(p-1); while(n%p==0) n/=p; } } if(n>1) ret=ret/n*(n-1); // 注意与后面的比较,时间更快,只有一个比sqrt(n)大的一个素数因子
return ret; } int main() { int n; while(scanf("%d",&n),n) printf("%d\n",Euler(n)); return 0; }
#include <iostream> #include<cmath> #include <algorithm> using namespace std; int Euler(int n) { int ret=n,p; for( p=2;p<=n;p++) { if(n%p==0) { ret=ret/p*(p-1); while(n%p==0) n/=p; } } if(n>1) ret=ret/p*(p-1); return ret; } int main() { int n; while(scanf("%d",&n),n) printf("%d\n",Euler(n)); return 0; }