UVA 11426 - GCD - Extreme (II) GCD性质例题

题目链接：
http://acm.hust.edu.cn/vjudge/contest/70017#problem/O
题意：
给定一个正整数N$(N<4000000)$， 求$G=\sum_{i=1}^{i<N}\sum_{j=i+1}^{j{\leq}N}gcd(i,j)$的值
分析
已知$f(N)=gcd(1,N) + gcd (2,N)+{\cdots}+gcd(N-1,N)$
即$j = N , f(N) = \sum_{i=1}^{i<N}gcd(i,N)$
则结果为$f(2) + f(3) + f(4) + {\cdots} + f(N)$
即$ans = \sum_{i=2}^{i{\leq}N}f(i)$
根据gcd性质，若$gcd(k,n)=i$,则必有$gcd(k/i,n/i)=1$，即$k/i与n/i$互质，
那么枚举$i$的值，$f(n)$等于$N/i$的欧拉函数值的和

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <cmath>
#include <set>
#define MOD 1000000007
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
/*---------------------------head files----------------------------------*/
const int maxn = 4000009 ;

int phi[maxn], prime[maxn / 10];
long long s[maxn],f[maxn];
int tot ;
bool not_prime[maxn];

void init()
{
    int i, j;
    phi[1] = 1;
    for ( int i = 2; i <= 4000000 ; i ++ )
    {
        if ( !not_prime[i] )
        {
            phi[i] = i - 1;
            prime[++tot] = i;
        }
        {
            for ( j = 1; j <= tot && i * prime[j] <= 4000000; j++ )
            {
                not_prime[i * prime[j]] = 1;
                if ( i % prime[j] == 0 )
                {
                    phi[i * prime[j]] = phi[i] * prime[j];
                    break;
                }
                else
                {
                    phi[i * prime[j]] = phi[i] * ( prime[j] - 1 );
                }
            }
        }
        //for ( i = 1; i <= 4000000; i++ ) sum[i] = ( sum[i - 1] + phi[i] ) % mod;
    }
}
//预处理素数与欧拉函数值
int main()
{
    init();
    for (int i = 2 ; i < maxn ; i ++)
    {
        for (int j = 1 ; j * i <maxn ; j ++ )
        {
            f[j*i] += phi[i]*j;
        }
    }
    for (int i = 2 ; i < maxn ; i ++)
    {
        s[i] = s[i-1] + f[i];
    }
    int n;
    while (~scanf("%d",&n)&&n)
    {
        printf("%lld\n",s[n]);
    }
}