题意：
给定(n,m,p)，求
[sum_{a=1}^nsum_{b=1}^mfrac{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
[egin{aligned}
sum_{a=1}^nsum_{b=1}^mfrac{varphi(ab)}{varphi(a)varphi(b)}mod p&=sum_{a=1}^nsum_{b=1}^mfrac{gcd(a,b)}{(p_1 - 1)(p_2-1)dots (p_k-1)}mod p\
&=sum_{a=1}^nsum_{b=1}^mfrac{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。
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
", ans);
}
return 0;
}