BZOJ2818 GCD

题意：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. N <= 10^7

http://www.lydsy.com/JudgeOnline/problem.php?id=2818

 

欧拉线性筛做法：

枚举每个素数p，对答案的贡献就是 phi[ ⌊ n / p ⌋ ] 的前缀和 * 2 - 1  (-1因为(1, 1)只计算一次)

#include<bits/stdc++.h>

using namespace std;

const int MAXN = 10000010;
int p[MAXN], phi[MAXN], x;
long long s[MAXN];
bool f[MAXN];
int n;

void getphi() {
    f[1] = 1;
    phi[1] = 1;
    for (int i = 2; i <= n; ++i) {
        if (!f[i]) {
            p[++x] = i;
            phi[i] = i - 1;
            for (int j = 2; j * i <= n; ++j) {
                f[i * j] = 1;
                if (j % i ==  0) phi[i * j] = phi[j] * i;
                else phi[i * j] = phi[j] * phi[i];
            }
        }
    }
}

int main() {
    x = 0;
    cin >> n;
    getphi();
    s[0] = 0;
    for (int i = 1; i <= n; ++i) s[i] = s[i - 1] + phi[i];
    long long ans = 0;
    for (int i = 1; i <= x; ++i) {
        ans += s[n / p[i]] * 2 - 1; 
    }
    cout << ans << "\n";
    return 0;
}

 

莫比乌斯反演做法：

第一次写反演emmm意外顺利

#include<bits/stdc++.h>

using namespace std;

const int MAXN = 10000010;
int p[MAXN], mob[MAXN], x;
long long s[MAXN];
bool f[MAXN];
int n;

void getmob() {
    f[1] = 1;
    mob[1] = 1;
    for (int i = 2; i <= n; ++i) {
        if (!f[i]) {
            p[++x] = i;
            mob[i] = -1;
            for (int j = 2; j * i <= n; ++j) {
                f[i * j] = 1;
                if (j % i ==  0) mob[i * j] = 0;
                else mob[i * j] = -mob[j];
            }
        }
    }
}

int main() {
    x = 0;
    cin >> n;
    getmob();
    s[0] = 0;
    for (int i = 1; i <= n; ++i) s[i] = s[i - 1] + mob[i];
    long long ans = 0;
    for (int d = 1; d <= x && p[d] <= n; ++d) {
        long long nx = n / p[d];
        long long j = 0;
        for (int i = 1; i <= nx; i = j + 1) {
            j = min(nx, nx / (nx / i));
            ans += (nx / i) * (nx / i) * (s[j] -s[i - 1]);            
        }
    }
    cout << ans << "\n";
    return 0;
}