UVA 11426 (欧拉函数)

白书125页例题。

gcd(x,n)所有可能的值是n的约数，要满足gcd(x,n) = i(i是n的约数)必然有gcd (x/i, n/i) = 1，也就是ph(n/i)的值。也就是对于每一个数i，如果i是n的约数，那么i在以n结尾的gcd中出现phi(i)次，这就是i对答案的贡献。

#include <bits/stdc++.h>
using namespace std;
#define maxn 4000000

long long phi[maxn+5];
long long f[maxn+5], sum[maxn+5];

void init () { //打表
    for (int i = 2; i <= maxn; i++)
        phi[i] = 0;
    phi[1] = 1;
    for (int i = 2; i <= maxn; i++) if (!phi[i]) {
        for (int j = i; j <= maxn; j += i) {
            if (!phi[j])
                phi[j] = j;
            phi[j] = phi[j] / i * (i-1);
        }
    }
    memset (f, 0, sizeof f);
    for (int i = 1; i <= maxn; i++) {
        for (int j = i+i; j <= maxn; j += i)
            f[j] += i*phi[j/i];
    } 
    sum[2] = 1;
    for (int i = 3; i <= maxn; i++)
        sum[i] = sum[i-1]+f[i];
    return ;
}

int n;

int main () {
    //freopen ("in", "r", stdin);
    init ();
    while (cin >> n && n) {
        cout << sum
 << endl;
    }
    return 0;
}