HDU 6390 GuGuFishtion（莫比乌斯反演 + 欧拉函数性质 + 积性函数）题解

Posted 2022-09-26 kirinsb

题意：

给定\(n,m,p\)，求
\[ \sum_a=1^n\sum_b=1^m\frac\varphi(ab)\varphi(a)\varphi(b)\mod p \]

思路：

由欧拉函数性质可得：\(x,y\)互质则\(\varphi(xy)=\varphi(x)\varphi(y)\)；\(p\)是质数则\(\varphi(p^a)=(p-1)^a-1\)。因此，由上述两条性质，我们可以吧\(a,b\)质因数分解得到
\[ \beginaligned \sum_a=1^n\sum_b=1^m\frac\varphi(ab)\varphi(a)\varphi(b)\mod p&=\sum_a=1^n\sum_b=1^m\fracgcd(a,b)(p_1 - 1)(p_2-1)\dots (p_k-1)\mod p\&=\sum_a=1^n\sum_b=1^m\fracgcd(a,b)\varphi(gcd(a,b))\mod p\&=\sum_k\sum_k|d\mu(\fracdk)F(d)*k*inv[\varphi(k)] \mod p \endaligned \]
有点卡常。

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 = 1e6 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 131;
const ll MOD = 1000000007;
using namespace std;

int mu[maxn], vis[maxn];
int prime[maxn], cnt, phi[maxn];
ll inv[maxn];
void init(int n)
    for(int i = 0; i <= n; i++) vis[i] = mu[i] = 0;
    cnt = 0;
    mu[1] = 1;
    phi[1] = 1;
    for(int i = 2; i <= n; i++) 
        if(!vis[i])
            prime[cnt++] = i;
            mu[i] = -1;
            phi[i] = i - 1;
        
        for(int j = 0; j < cnt && prime[j] * i <= n; j++)
            vis[prime[j] * i] = 1;
            if(i % prime[j] == 0)
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            
            mu[i * prime[j]] = -mu[i];
            phi[i * prime[j]] = phi[i] * (prime[j] - 1);
        
    

void init2(int n, ll p)
    inv[0] = inv[1] = 1;
    for(int i = 2; i <= n; i++)
        inv[i] = (p - p / i) * inv[p % i] % p;


int main()
    init(1e6);
    int T;
    scanf("%d", &T);
    while(T--)
        ll n, m, p;
        scanf("%lld%lld%lld", &n, &m, &p);
        ll mm = min(n, m);
        init2(mm, p);
        ll ans = 0;
        for(int k = 1; k <= mm; k++)
            ll temp = 0;
            for(int d = k; d <= mm; d += k)
                temp += 1LL * mu[d / k] * (n / d) * (m / d);
            
            temp = temp * k % p * inv[phi[k]] % p;
            ans = (ans + temp) % p;
        
        ans = (ans + p) % p;
        printf("%lld\n", ans);
    
    return 0;