LightOj 1375(欧拉变形)

lightoj 1375
题目大意：
求小于$n$ 的数字中，任意两个不相等的数字的$lcm$和，即$\displaystyle\sum_{i=1,j=1,i \neq j}^{i=n,j=n}{lcm(i,j)}$;
思路：
$lcm(1,2)$  $lcm(1, 3)$ $lcm(1,4)$ … $lcm(1,n)$
                  $lcm(2,3)$ $lcm(2, 4)$ … $lcm(2,n)$
                                   $lcm(3,4)$ … $lcm(3,n)$
                                                     …$lcm(n-1,n)$
即可求$\displaystyle\sum_{i=2}^n{\displaystyle\sum_{j=1}^{j=i-1}{lcm(j,i)}}$;
$lcm(i, n) = \displaystyle{{i * n} \over {gcd(i,n)}}$, 令$gcd(i, n) = G;$
$lcm(i,n) = {i\over g} * {n \over g} * g;$ $gcd({i\over g},{{n\over g})} = 1$;
又已知小于$n$且与$n$互质的数的和等于$\displaystyle{n* \varphi(n)}\over{2}$, 证明;
$\displaystyle\sum_{i=1}^{n-1}{lcm(i,n)} = {{{i \over g}* \varphi({n \over g})}\over{2}}*g$;
即: $\displaystyle\sum_{i=1}^{n-1}{lcm(i,n)} = \displaystyle\sum_{d \mid n, d \neq n}{{d * \varphi (d)}\over2}*n$; 即枚举n的因子；

取模$2^{64}$, 变量定义为$unsigned$ $long$ $long$; 溢出$2^{64}$即为取模$2^{64}$;

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#define N 3000001
#define LL unsigned long long

using namespace std;

int euler[N];
LL f[N], a[N];

void init_Euler()
{
    euler[1] = 1;

    for (int i = 2; i < N; i++)
    {
        euler[i] = i;
    }

    for (int i = 2; i < N; i++)
    {
        if (euler[i] == i)
        {
            for (int j = i; j < N; j += i)
            {
                euler[j] = euler[j] / i * (i - 1);
            }
        }
    }

    memset(a, 0, sizeof(a));  
    memset(f, 0, sizeof(f)); 

    for(int i = 2; i < N; i++)  
    {  
        for(int j = i; j < N; j += i)  
        {  
            f[j] += ((LL)euler[i]) * i / 2 * j;  
        }  

        a[i] = a[i - 1] + f[i];  
    }
}

int main()
{
    init_Euler();

    int T;
    scanf("%d", &T);

    for (int cas = 1; cas <= T; cas++)
    {
        int n;
        scanf("%d", &n);

        printf("Case %d: %llu\n", cas, a[n]);
    }
    return 0;
}