HDU 4746 Mophues（莫比乌斯反演）题解

题意：

\(Q\leq5000\)次询问，每次问你有多少对\((x,y)\)满足\(x\in[1,n],y\in[1,m]\)且\(gcd(x,y)\)的质因数分解个数小于等于\(p\)。\(n,m,p\leq5e5\)。

思路：

题目即求

\[\sum_{k}\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=k]\quad,k满足质因数个数\leq p
\]

令\(f(n)\)为\(gcd\)为\(n\)的对数，\(F(n)\)为\(gcd\)为\(n\)倍数的对数。

由莫比乌斯反演可得

\[\begin{aligned}
\sum_{k}\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=k]
&=\sum_kf(k)\\
&=\sum_k\sum_{k|d}\mu(\frac{d}{k})F(d)\\
&=\sum_k\sum_{k|d}\mu(\frac{d}{k})\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor\\
&=\sum_d\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor\sum_{k|d}\mu(\frac{d}{k})
\end{aligned}
\]

\(\sum_{k|d}\mu(\frac{d}{k})\)可以直接打表打出来。

直接枚举\(d\)，因为\(\lfloor\frac{n}{k}\rfloor\lfloor\frac{m}{k}\rfloor\)很多都是重复的，那么我可以直接分块加速，先求\(\sum_{k|d}\mu(\frac{d}{k})\)的前缀和，然后每次选\(i\)~\(min(\lfloor \frac{n}{\lfloor \frac{n}{i}\rfloor}\rfloor,\lfloor \frac{m}{\lfloor \frac{m}{i}\rfloor}\rfloor)\)这个区间走，那么\(\sqrt{(min(n, m))}\)就遍历完了。

代码：

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 5e5 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 131;
const ll MOD = 1e9;
using namespace std;
int num[maxn], sum[21][maxn];
int mu[maxn], vis[maxn];
int prime[maxn], cnt;
void getmu(int n){
	memset(vis, 0, sizeof(vis));
	memset(mu, 0, sizeof(mu));
	memset(num, 0, sizeof(num));
	cnt = 0;
	mu[1] = 1;
	for(int i = 2; i <= n; i++) {
		if(!vis[i]){
            prime[cnt++] = i;
            mu[i] = -1;
            num[i] = 1;
        }
		for(int j = 0; j < cnt && prime[j] * i <= n; j++){
			vis[i * prime[j]] = 1;
			num[i * prime[j]] = num[i] + 1;
			if(i % prime[j] == 0) break;
			mu[i * prime[j]] = -mu[i];
		}
	}
}

void init(){
    memset(sum, 0, sizeof(sum));    //sum[p][d]:d的除数的质因子个数为p的sum(mu)
    for(int i = 1; i <= 5e5; i++){
        for(int j = i; j <= 5e5; j += i){
            sum[num[i]][j] += mu[j / i];
        }
    }
    for(int i = 1; i <= 5e5; i++){  //d的除数质因子个数小于p的sum(mu)
        for(int j = 1; j <= 19; j++){
            sum[j][i] += sum[j - 1][i];
        }
    }

    for(int i = 1; i <= 5e5; i++){
        for(int j = 0; j <= 19; j++){
            sum[j][i] += sum[j][i - 1];
        }
    }
}
int main(){
    getmu(5e5);
    init();
    ll n, m;
    int p, T;
    scanf("%d", &T);
    while(T--){
        ll ans = 0;
        scanf("%lld%lld%d", &n, &m, &p);
        if(p > 19){
            printf("%lld\n", n * m);
            continue;
        }
        for(int i = 1; i <= min(n, m);){
            int l, r;
            l = i, r = min(n / (n / i), m / (m / i));
            ans += 1LL * (n / i) * (m / i) * (sum[p][r] - sum[p][l - 1]);
            i = r + 1;
        }
        printf("%lld\n", ans);
    }
    return 0;
}