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
\[ \begin{aligned} \sum_{a=1}^n\sum_{b=1}^m\frac{\varphi(ab)}{\varphi(a)\varphi(b)}\mod p&=\sum_{a=1}^n\sum_{b=1}^m\frac{gcd(a,b)}{(p_1 - 1)(p_2-1)\dots (p_k-1)}\mod p\\ &=\sum_{a=1}^n\sum_{b=1}^m\frac{gcd(a,b)}{\varphi(gcd(a,b))}\mod p\\ &=\sum_{k}\sum_{k|d}\mu(\frac{d}{k})F(d)*k*inv[\varphi(k)] \mod p \end{aligned} \]
有点卡常。
#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; } 来源：博客园
作者：KirinSB
链接：https://www.cnblogs.com/KirinSB/p/11451655.html