bzoj3309 - DZY Loves Math（莫比乌斯函数，类积性函数筛法）

题目

对于正整数n，定义f(n)为n所含质因子的最大幂指数。例如f(12250)=f(2^1 * 5^3 * 7^2)=3, f(10007)=1, f(1)=0。
给定正整数a,b，求\(\sum\limits_{i=1}^{a}{\sum\limits_{j=1}^{b}{f(\gcd(i,j))}}\)。

题解

\[\sum\limits_{i=1}^{a}{\sum\limits_{j=1}^{b}{f(\gcd(i,j))}} \]

\[\sum\limits_d\sum\limits_{i=1}^{\frac{a}{d}}{\sum\limits_{j=1}^{\frac{b}{d}}{f(d)[\gcd(i,j)=1]}} \]

\[\sum\limits_d\sum\limits_{i=1}^{\frac{a}{d}}{\sum\limits_{j=1}^{\frac{b}{d}}{f(d)\sum\limits_{d_1|gcd(i,j)}{\mu(d_1)}}} \]

\[\sum\limits_d{\sum\limits_{d_1}{\mu(d_1)f(d)\lfloor\frac{a}{dd_1}\rfloor\lfloor\frac{b}{dd_1}\rfloor}} \]

令\(T=dd_1\)

\[\sum\limits_T^{min(a, b)}{\lfloor\frac{a}{T}\rfloor\lfloor\frac{b}{T}\rfloor\sum\limits_{d|T}{f(d)\mu(\frac{T}{d})}} \]

现在关键是求\(\sum\limits_{d|T}{f(d)\mu(\frac{T}{d})}\)里的\(f\)。观察可发现，\(f\)有类似积性函数的性质，即当\(\gcd(i,j)=1\)时，有\(f(ij)=max(f(i),f(j))\)，可以使用线性筛。多维护一个最小质因子的次数的函数num，在i%p==0情况时容易想到有\(f(i*p)=\max(f(i), num(i *p))\)

然后就是常规操作数论分块了。

#include <bits/stdc++.h>

#define endl '\n'
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define mp make_pair
#define seteps(N) fixed << setprecision(N) 
typedef long long ll;

using namespace std;
/*-----------------------------------------------------------------*/

ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
#define INF 0x3f3f3f3f

const int N = 1e7 + 10;
const double eps = 1e-5;


int mu[N];
ll f[N];
int prime[N];
int cnt;
int vis[N];
int num[N];
int mx[N];

void init() {
    mu[1] = num[1] = 1;
    for(int i = 2; i < N; i++) {
        if(!vis[i]) {
            mu[i] = -1;
            num[i] = 1;
            mx[i] = 1;
            prime[cnt++] = i;
        }
        for(int j = 0; j < cnt; j++) {
            int p = prime[j];
            if(i * p > N) break;
            vis[i * p] = 1;
            if(i % p == 0) {
                mu[i * p] = 0;
                num[i * p] = num[i] + 1;
                mx[i * p] = max(num[i * p], mx[i]); 
                break;
            }
            mu[i * p] = -mu[i];
            num[i * p] = 1;
            mx[i * p] = mx[i];
        }
    }
    num[1] = 0;
    for(int i = 1; i < N; i++) {
        for(int j = i; j < N; j += i) {
            f[j] += mx[i] * mu[j / i];
        }
    }
    for(int i = 2; i < N; i++) f[i] += f[i - 1];
}


int main() {
    IOS;
    init();
    int t;
    cin >> t;
    while(t--) {
        ll ans = 0;
        int n, m;
        cin >> n >> m;
        int cur = 1;
        while(cur <= min(n, m)) {
            int p1 = n / (n / cur);
            int p2 = m / (m / cur);
            int nt = min(min(n, m), min(p1, p2));
            ans += 1ll * (n / cur) * (m / cur) * (f[nt] - f[cur - 1]);
            cur = nt + 1;
        }
        cout << ans << endl;

    }
}