这题主要是求1..n-1中与n互质的数的个数，并且要将其累加起来

我写的代码套用了求欧拉数的函数，结果超时，求大牛点拨优化方法

#include <iostream> #include <cstdio> using namespace std; int eun[1000005] = {0}; //求1..n-1中与n互质的数的个数 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,i,sum; for (i = 2;i <= 1000000 ;i++) { eun[i] = eular(i); } while (1) { //cin>>n; scanf("%d",&n); if (n == 0) break; sum = 0; for (i = 2;i <= n ;i++) { sum += eun[i]; } //cout<<sum<<endl; printf("%d",sum); } return 0; } 

 

在网上看到网友的解答，摘录余下：

#include <stdio.h> #include <string.h> #include <math.h> const int maxn = 1000000+1; //const int maxn = 1000+5; double save[maxn]; void init() { int i,j; for (i = 2; i < maxn; i ++) save[i] = i; for (i = 2; i < maxn; i ++) { if(save[i] == i) { save[i] = i-1; for(j=2*i;j<maxn;j+=i) save[j]=save[j]*(1-(double)1/(double)i); } } for (i = 2; i < maxn; i ++) save[i] += save[i-1]; } int main() { int n; init(); while (scanf("%d",&n),n != 0) { printf("%.0lf/n",save[n]); } return 0; }