Luogu p2568 GCD

题目链接

https://www.luogu.com.cn/problem/P2398

问题分析：

代码如下：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 1e7 + 10;
int phi[maxn];
int primes[maxn];
ll sum[maxn];
bool isnp[maxn];
int cnt = 0;
void init(int n)
{
        phi[1] = 1;
        for (int i = 2; i <= n; i++)
        {
                if (!isnp[i])
                        primes[++cnt] = i, phi[i] = i - 1; 
                        // 性质一，指数为1的情形
                for (int p = 1; p <= cnt && primes[p] * i <= n; p++)
                {
                        isnp[primes[p] * i] = 1;
                        if (i % primes[p] == 0)
                        {
                                phi[primes[p] * i] = phi[i] * primes[p]; 
                                // 性质二
                                break;
                        }
                        else
                                phi[primes[p] * i] = phi[primes[p]] * phi[i]; // 这时肯定互质，用性质三
                }
        }
        for (int i = 1; i <= n; i++)
                sum[i] = sum[i - 1] + phi[i];
}
//前缀数组(用来储存gcd(i,j) == 1的素数对)
int main()
{
        ll n;
        cin >> n;
        init(n);
        ll ans = 0;
        for (ll i = 1; i <= cnt; i++)
        {
                ans = ans + (sum[n / primes[i]]) * 2 - 1;
        }
        cout << ans << endl;
        return 0;
}