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給定整數N，求1<=x,y<=N且Gcd(x,y)為素數的數對(x,y)有多少對.

1<=N<=10^7

(1)莫比烏斯反演法

發現就是YY的GCD，左轉YY的GCD粘過來就行

代碼太醜，沒開O2 TLE5個點

#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; //別忘了初始化。。。
        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\n", ans);
    // }
    return 0;
}
 

(2)歐拉函數法

對於一個\(p\)我們發現\(\sum_{i=1}^n\sum_{j=1}^n[\gcd(i,j)=p]\)即為\(\sum_{i=1}^{n/p}\sum_{j=1}^{n/p}[\gcd(i,j)=1]\)

左轉SDOI儀仗隊那題，發現這個式子就是\(2\varphi(\lfloor\frac n p\rfloor)+1\)

線性篩就行

(一個月前的代碼

#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;
}
 

luogu2658 GCD(莫比烏斯反演/歐拉函數)