[HDU1787]GCD Again（根n求phi）

题目描述

题解

代码

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define LL long long

LL T,n,ans;

inline LL phi(LL x){
LL ans=x;
for (LL i=2;i*i<=x;++i)
if (x%i==0){
ans=ans*(i-1)/i;
while (x%i==0) x/=i;
}
if (x>1) ans=ans*(x-1)/x;
return ans;
}

int main(){
while (~scanf("%lld",&n)){
if (!n) break;
ans=n-1-phi(n);
printf("%lld\n",ans);
}
}