Description#

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.

Input#

一个整数N

Output#

如题

Sample Input#

4

Sample Output#

4

题解#

$\sum_{d}{\sum_{i=1}^{n}{\sum_{j=1}^{n}[gcd(i,j)=d]}}$

$= \sum_{d}{\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}{\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}{[gcd(i,j)=1]}}}$

$= \sum_{T}{\lfloor\frac{n}{T}\rfloor \lfloor\frac{n}{T}\rfloor\sum_{d|T}{\mu(\frac{n}{T})}}$
令 $f[i]=\sum_{d|T}^{} \mu(\frac{T}{d})$

然后线性筛

#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 10000000;
int mu[N + 5], prime[N + 5], g[N + 5];
long long sum[N + 5], cnt;
bool isnprime[N + 5];
void get_g()
{
    mu[1] = 1;
    for (int i = 2; i <= N; i++)
    {
        if (!isnprime[i])
        {
            prime[++cnt] = i;
            mu[i] = -1;
            g[i] = 1;
        }
        for (int j = 1; j <= cnt && i * prime[j] <= N; j++)
        {
            isnprime[i * prime[j]] = 1;
            if (i % prime[j])
                mu[i * prime[j]] = -mu[i], g[i * prime[j]] = mu[i] - g[i];
            else
            {
                mu[i * prime[j]] = 0;
                g[i * prime[j]] = mu[i];
                break;
            }
        }
    }
    for (int i = 1; i <= N; i++)
    {
        sum[i] = sum[i - 1] + g[i];
    }
}
int main()
{
    get_g();
    int n;
    scanf("%d", &n);
    long long ans = 0, last;
    for (int i = 1; i <= n; i = last + 1)
    {
        last = n / (n / i);
        ans += (long long)(sum[last] - sum[i - 1]) * (n / i) * (n / i);
    }
    printf("%lld\n", ans);
}