luogu2658 GCD(莫比乌斯反演/欧拉函数)

Posted 2021-02-04 oier

link

给定整数N，求1<=x,y<=N且Gcd(x,y)为素数的数对(x,y)有多少对.

1<=N<=10^7

(1)莫比乌斯反演法

发现就是YY的GCD，左转YY的GCD粘过来就行

代码太丑，没开O2 TLE5个点

#include <cstdio>
#include <functional>
using namespace std;

const int fuck = 10000000;
int prime[10000010], tot;
bool vis[10000010];
int mu[10000010], sum[10000010];

int main()
{
    mu[1] = 1;
    for (int i = 2; i <= fuck; i++)
    {
        if (vis[i] == false) prime[++tot] = i, mu[i] = -1;
        for (int j = 1; j <= tot && i * prime[j] <= fuck; j++)
        {
            vis[i * prime[j]] = true;
            if (i % prime[j] == 0) break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for (int i = 1; i <= tot; i++)
        for (int j = 1; j * prime[i] <= fuck; j++)
            sum[j * prime[i]] += mu[j];
    for (int i = 1; i <= fuck; i++)
        sum[i] += sum[i - 1];
    // int t; scanf("%d", &t);
    // while (t --> 0)
    // {
        int n, m;
        long long ans = 0; //别忘了初始化。。。
        scanf("%d", &n), m = n;
        if (n > m) {int t = m; m = n; n = t; }
        for (int i = 1, j; i <= n; i = j + 1)
        {
            j = min(n / (n / i), m / (m / i));
            ans += (sum[j] - sum[i - 1]) * (long long)(n / i) * (m / i);
        }
        printf("%lld
", ans);
    // }
    return 0;
}

(2)欧拉函数法

对于一个(p)我们发现(sum_{i=1}^nsum_{j=1}^n[gcd(i,j)=p])即为(sum_{i=1}^{n/p}sum_{j=1}^{n/p}[gcd(i,j)=1])

左转SDOI仪仗队那题，发现这个式子就是(2varphi(lfloorfrac n p floor)+1)

线性筛就行

(一个月前的代码

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

int vis[10000010];
long long phi[10000010];
int prime[1000010], tot, n;

int main()
{
    cin >> n;
    phi[1] = 1;
    for (int i = 2; i <= n; i++)
    {
        if (vis[i] == 0)
            prime[++tot] = i, phi[i] = i - 1;
        for (int j = 1; j <= tot && i * prime[j] <= n; j++)
        {
            vis[i * prime[j]] = true;
            if (i % prime[j] == 0)
            {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            phi[i * prime[j]] = phi[i] * (prime[j] - 1);
        }
        vis[i] ^= 1;
        vis[i] += vis[i - 1];
        phi[i] += phi[i - 1];
    }
    long long ans = 0;
    for (int i = 1; i <= tot; i++)
        ans += 2 *  phi[n / prime[i]] - 1;
    cout << ans << endl;
    return 0;
}