Relatives
| Time Limit:1s | Memory limit:32M |
| Accepted Submit:387 | Total Submit:1304 |
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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. 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. For each test case there should be single line of output answering the question posed above. Sample Input
Sample Output
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解题：
题目要求1..n-1中与n互质的数的个数，运用欧拉函数求解。
假设要求x的欧拉函数值，则可以按照下面求解。
举例：φ(72)=φ(2^3×3^2)=(2-1)2^(3-1)×(3-1)3^(2-1)=24
代码实现的时候，遇到是素数p1的话，先乘以（p1-1），再计算是p1的几次幂，然后乘起来；继续计算下个素数。
欧拉函数是积性函数——若m,n互质，φ(mn)=φ(m)φ(n)。
#include <iostream> using namespace std; int eular(int n) { int ret=1,i; for (i=2;i*i<=n;i++) if (n%i==0) { n/=i,ret*=i-1; while (n%i==0) n/=i,ret*=i; } if (n>1) ret*=n-1; return ret; } int main() { int n; while (cin>>n && n!=0) { cout<<eular(n)<<endl; } return 0; }
