洛谷 P4450双亲数

题意：

给定$A, B, d$, 求$\sum_{a=1}^{A}\sum_{b=1}^{B}[\gcd(a, b) = d]$

思路：

$\sum_{a=1}^{A}\sum_{b=1}^{B}[\gcd(a, b) = d]$

简化式子：

$\sum_{a=1}^{\lfloor\frac{A}{d}\rfloor}\sum_{b=1}^{\lfloor\frac{B}{d}\rfloor}[\gcd(a, b)=1]$

将$[\gcd(a, b) = 1]$替换为$\varepsilon(\gcd(a, b))$:

$\sum_{x=1}^{\lfloor\frac{A}{d}\rfloor}\sum_{y=1}^{\lfloor\frac{B}{d}\rfloor}\varepsilon(\gcd(a, b))$

展开$\varepsilon$函数得到：

$\sum_{x=1}^{\lfloor\frac{A}{d}\rfloor}\sum_{y=1}^{\lfloor\frac{B}{d}\rfloor}\sum_{k|\gcd(a, b)} \mu(k)$

变换求和顺序，先枚举$k|\gcd(a, b)$得：

$\sum_{k=1}\mu(k)\sum_{x=1}^{\lfloor\frac{A}{d}\rfloor}[k|x]\sum_{y=1}^{\lfloor\frac{B}{d}\rfloor}[k|y]$

已知$1~\lfloor\frac{A}{d}\rfloor$中$k$的倍数有$\lfloor\frac{A}{kd}\rfloor$个：

$\sum_{k=1}\mu(k)\lfloor\frac{A}{kd}\rfloor\lfloor\frac{B}{kd}\rfloor$

Code:

#pragma GCC optimize(3)
#pragma GCC optimize(2)
#include <map>
#include <set>
// #include <array>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
// #include <unordered_map>

using namespace std;

typedef long long ll;
typedef pair<int, int> PII;

#define Time (double)clock() / CLOCKS_PER_SEC

#define sd(a) scanf("%d", &a)
#define sdd(a, b) scanf("%d%d", &a, &b)
#define slld(a) scanf("%lld", &a)
#define slldd(a, b) scanf("%lld%lld", &a, &b)

const int N = 2e6 + 20;
const int M = 1e5 + 20;
const int mod = 1e9 + 7;
const double eps = 1e-6;

int cnt = 0, primes[N], mu[N], sum[N];
bool st[N];

void get(int n){
    mu[1] = 1;
    for(int i = 2; i <= n; i ++){
        if(!st[i]){
            primes[cnt ++] = i;
            mu[i] = -1;
        }

        for(int j = 0; primes[j] <= n / i; j ++){
            st[i * primes[j]] = true;
            if(i % primes[j] == 0){
                mu[i * primes[j]] = 0;
                break;
            }
            mu[i * primes[j]] = -mu[i];
        }
    }
    sum[0] = 0;
    for(int i = 1; i <= n; i ++){
        sum[i] = sum[i - 1] + mu[i];
    }

}

void solve()
{
    int a, b, d;
    sdd(a, b);
    sd(d);

    ll ans = 0;
    a /= d, b /= d;

    for(int k = 1, r; k <= min(a, b); k = r + 1){
        r = min(a / (a / k), b / (b / k));
        ans = ans + (ll)(sum[r] - sum[k - 1]) * (a / k) * (b / k); 
    }
    printf("%lld\n", ans);
}


int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("/home/jungu/code/in.txt", "r", stdin);
    // freopen("/home/jungu/code/out.txt", "w", stdout);
    // freopen("/home/jungu/code/practice/out.txt","w",stdout);
#endif
    // ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    // sd(T);
    get(2000000);
    // init();
    // int cas = 1;
    while (T--)
    {
        // printf("Case #%d:", cas++);
        solve();
    }

    return 0;
}