GCD-Extreme (II) (UVA-11426, Euler function)

1. Title link:

UVA-11426

2. The main idea of the topic:

$\sum_{i = 1}^{N}\sum_{j = i +1}^{N}gcd(i, j)$

3. Analysis:

Posting a solution to the problem of Juju is naturally not something I can think of...

Another thing that has bothered me for a long time is how to calculate F[n] quickly.

As written in the Juju blog $F[n]=\sum i\phi(\frac{n}{i})$

Ben Konjac only thought of $O (N ^ 2)$ the complexity calculation F[], and was crying stupidly by myself _(:3」∠)_

The positive solution is to calculate the contribution to F[] when gcd(x, j) = i. It is easy to get i | j, and the complexity is $O (NlnN)$

4. Code implementation:

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

typedef long long ll;

const int M = (int)4e6;

bool is_prime[M + 5];
int cnt, prime[M + 5];
int phi[M + 5];
ll sum[M + 5];

void init()
{
    cnt = 0;
    memset(is_prime, 1, sizeof(is_prime));
    is_prime[0] = is_prime[1] = 0;
    for(int i = 2; i <= M; ++i)
    {
        if(is_prime[i])
        {
            prime[++cnt] = i;
            phi[i] = i - 1;
        }
        for(int j = 1; j <= cnt && i * prime[j] <= M; ++j)
        {
            is_prime[i * prime[j]] = 0;
            if(i % prime[j] == 0)
            {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            phi[i * prime[j]] = phi[i] * (prime[j] - 1);
        }
    }
}

int main()
{
    init();
    for(int i = 1; i <= M; ++i)
    {
        for(int j = 2 * i; j <= M; j += i)
        {
            sum[j] += 1ll * i * phi[j / i];
        }
    }
    for(int i = 1; i <= M; ++i)
        sum[i] += sum[i - 1];
    int n;
    while(~scanf("%d", &n) && n)
    {
        printf("%lld\n", sum[n]);
    }
    return 0;
}