推式子
[sum_{i=1}^{n} sum_{j=1}^{m} [gcd(i,j) = d]\
sum_{i=1}^{frac nd} sum_{j=1}^{frac md} [gcd(i,j) = 1] \
----------------\
sum_{i=1}^{n} sum_{j=1}^{m} [gcd(i,j) = 1]\
sum_{i=1}^{n} sum_{j=1}^{m} sum_{p | i, p|j} mu(p)\
sum_{p=1}^{n} mu(p) sum_{i=1}^{frac np} sum_{j=1}^{frac mp} \
]
整除分块即可
code
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5;
int p[N + 5], prime[N / 10], cnt;
int mu[N + 5];
ll f(int n){
return 1ll * n;
}
ll g(ll n, ll m){
ll ret = 0;
for(int l = 1, r; l <= n; l = r + 1){
r = min(n, min(n/ (n/l), m/ (m/l)));
ret = ret + 1ll * (mu[r] - mu[l - 1]) * f(n/l) * f(m/l);
}
return ret;
}
int main(){
mu[1] = 1; p[0] = p[1] = 1;
for(int i = 2; i <= N; ++ i){
if(!p[i]) { prime[++ cnt] = i; mu[i] = -1; }
for(int j = 1; j <= cnt && 1ll * prime[j] * i <= N; ++ j){
p[prime[j] * i] = 1;
if(i % prime[j] == 0) break;
mu[prime[j] * i] = -mu[i];
}
}
for(int i = 1; i <= N; ++ i) mu[i] += mu[i - 1];
int T; scanf("%d",&T);
while(T --){
int a,b,d; scanf("%d%d%d",&a,&b,&d);
if(a > b) swap(a,b); a /= d; b /= d;
printf("%lld
",g(a, b));
}
return 0;
}