[SDOI2015] 约数个数和 - 莫比乌斯反演,整除分块

标签：image   分块   --   lang   line   lazy   ==   sum   int   

Description

设 \(d(x)\) 为 \(x\) 的约数个数，给定 \(n,m\)，求 \(\sum_{i=1}^n\sum_{j=1}^md(ij)\)，\(n,m,T \le 5\times 10^4\)

Solution

重要结论如黑色式子所示，其余部分为套路性的反演推导

#include <bits/stdc++.h>
using namespace std;

#define int long long
const int N = 50005;
const int MAXN = 50005;

bool isNotPrime[MAXN + 1];
int mu[MAXN + 1], phi[MAXN + 1], primes[MAXN + 1], cnt, h[MAXN+1];
inline void euler() {
    isNotPrime[0] = isNotPrime[1] = true;
    mu[1] = 1;
    phi[1] = 1;
    for (int i = 2; i <= MAXN; i++) {
        if (!isNotPrime[i]) {
            primes[++cnt] = i;
            mu[i] = -1;
            phi[i] = i - 1;
        }
        for (int j = 1; j <= cnt; j++) {
            int t = i * primes[j];
            if (t > MAXN) break;
            isNotPrime[t] = true;
            if (i % primes[j] == 0) {
                mu[t] = 0;
                phi[t] = phi[i] * primes[j];
                break;
            } else {
                mu[t] = -mu[i];
                phi[t] = phi[i] * (primes[j] - 1);
            }
        }
    }
    for(int i=1;i<=MAXN;i++) mu[i]+=mu[i-1];
}

int calch(signed x) {
    signed l=1;
    int ans=0;
    while(l<=x) {
        signed r=x/(x/l);
        ans+=1ll*(r-l+1)*(x/l);
        l=r+1;
    }
    return ans;
}

int solve(int n,int m) {
    if(n==0 || m==0) return 0;
    int ans=0,l=1,r=0;
    if(n>m) swap(n,m);
    while(l<=n) {
        r=min(n/(n/l),m/(m/l));
        ans+=(mu[r]-mu[l-1])*h[n/l]*h[m/l];
        l=r+1;
    }
    return ans;
}

signed main() {
    euler();
    int t,a,b,c,d,k;
    ios::sync_with_stdio(false);
    cin>>t;
    for(int i=1;i<=5e4;i++) h[i]=calch(i);
    while(t--) {
        int n,m;
        cin>>n>>m;
        cout<<solve(n,m)<<endl;
    }
}

[SDOI2015] 约数个数和 - 莫比乌斯反演,整除分块

标签：image   分块   --   lang   line   lazy   ==   sum   int   

原文地址：https://www.cnblogs.com/mollnn/p/13155293.html