gate

\(\large \sum\limits_{i=1}^{a}\sum\limits_{j=1}^{b}[gcd(i,j)=x]\)
\(=\large \sum\limits_{i=1}^{\frac{a}{x}}\sum\limits_{j=1}^{\frac{b}{x}}[gcd(i,j)=1]\)
\(=\large \sum\limits_{i=1}^{\frac{a}{x}}\sum\limits_{j=1}^{\frac{b}{x}} \sum\limits_{d|i,d|j}\mu(d)\)
\(=\large \sum\limits_{d}\mu(d)\sum\limits_{i=1}^{\frac{a}{xd}}\sum\limits_{j=1}^{\frac{b}{xd}}\)

預處理出\(\mu(i)\)的字首和，
求\(\large \lfloor\frac{a}{xd}\rfloor\lfloor\frac{b}{xd}\rfloor\)可以整除分塊，最多有\(2\sqrt{a} + 2\sqrt{b}\)種取值。
注意\((a/i)*(b/i)\)不要忘記打括號！因為是整除...

code

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#define MogeKo qwq
using namespace std;

const int maxn = 5e4+10;
const int N = 5e4;

int n,prime[maxn],mu[maxn],cnt;
long long a,b,d,f[maxn];
bool vis[maxn];

void Prime() {
	f[1] = mu[1] = 1;
	for(int i = 2; i <= N; i++) {
		if(!vis[i]) {
			prime[++cnt] = i;
			mu[i] = -1;
		}
		for(int j = 1; j <= cnt && i*prime[j] <= N; j++) {
			vis[i*prime[j]] = true;
			if(i % prime[j])
				mu[i*prime[j]] = -mu[i];
			else {
				mu[i*prime[j]] = 0;
				break;
			}
		}
		f[i] = f[i-1]+mu[i];
	}
}

long long solve(long long a,long long b) {
	long long ans = 0;
	if(a > b) swap(a,b);
	for(long long i = 1,r; i <= a; i = r+1) {
		r = min(a/(a/i),b/(b/i));
		ans += (f[r]-f[i-1]) * (a/i) * (b/i);
	}
	return ans;
}

int main() {
	Prime();
	scanf("%d",&n);
	while(n--) {
		scanf("%lld%lld%lld",&a,&b,&d);
		printf("%lld\n",solve(a/d,b/d));
	}
	return 0;
}