题意：
(Qleq5000)次询问，每次问你有多少对((x,y))满足(xin[1,n],yin[1,m])且(gcd(x,y))的质因数分解个数小于等于(p)。(n,m,pleq5e5)。
思路：
题目即求
[sum_{k}sum_{i=1}^nsum_{j=1}^m[gcd(i,j)=k]quad,k满足质因数个数leq p
]
令(f(n))为(gcd)为(n)的对数，(F(n))为(gcd)为(n)倍数的对数。
由莫比乌斯反演可得
[egin{aligned}
sum_{k}sum_{i=1}^nsum_{j=1}^m[gcd(i,j)=k]
&=sum_kf(k)\
&=sum_ksum_{k|d}mu(frac{d}{k})F(d)\
&=sum_ksum_{k|d}mu(frac{d}{k})lfloorfrac{n}{d}
floorlfloorfrac{m}{d}
floor\
&=sum_dlfloorfrac{n}{d}
floorlfloorfrac{m}{d}
floorsum_{k|d}mu(frac{d}{k})
end{aligned}
]
(sum_{k|d}mu(frac{d}{k}))可以直接打表打出来。
直接枚举(d)，因为(lfloorfrac{n}{k}
floorlfloorfrac{m}{k}
floor)很多都是重复的，那么我可以直接分块加速，先求(sum_{k|d}mu(frac{d}{k}))的前缀和，然后每次选(i)~(min(lfloor frac{n}{lfloor frac{n}{i}
floor}
floor,lfloor frac{m}{lfloor frac{m}{i}
floor}
floor))这个区间走，那么(sqrt{(min(n, m))})就遍历完了。
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 = 5e5 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 131;
const ll MOD = 1e9;
using namespace std;
int num[maxn], sum[21][maxn];
int mu[maxn], vis[maxn];
int prime[maxn], cnt;
void getmu(int n){
memset(vis, 0, sizeof(vis));
memset(mu, 0, sizeof(mu));
memset(num, 0, sizeof(num));
cnt = 0;
mu[1] = 1;
for(int i = 2; i <= n; i++) {
if(!vis[i]){
prime[cnt++] = i;
mu[i] = -1;
num[i] = 1;
}
for(int j = 0; j < cnt && prime[j] * i <= n; j++){
vis[i * prime[j]] = 1;
num[i * prime[j]] = num[i] + 1;
if(i % prime[j] == 0) break;
mu[i * prime[j]] = -mu[i];
}
}
}
void init(){
memset(sum, 0, sizeof(sum)); //sum[p][d]:d的除数的质因子个数为p的sum(mu)
for(int i = 1; i <= 5e5; i++){
for(int j = i; j <= 5e5; j += i){
sum[num[i]][j] += mu[j / i];
}
}
for(int i = 1; i <= 5e5; i++){ //d的除数质因子个数小于p的sum(mu)
for(int j = 1; j <= 19; j++){
sum[j][i] += sum[j - 1][i];
}
}
for(int i = 1; i <= 5e5; i++){
for(int j = 0; j <= 19; j++){
sum[j][i] += sum[j][i - 1];
}
}
}
int main(){
getmu(5e5);
init();
ll n, m;
int p, T;
scanf("%d", &T);
while(T--){
ll ans = 0;
scanf("%lld%lld%d", &n, &m, &p);
if(p > 19){
printf("%lld
", n * m);
continue;
}
for(int i = 1; i <= min(n, m);){
int l, r;
l = i, r = min(n / (n / i), m / (m / i));
ans += 1LL * (n / i) * (m / i) * (sum[p][r] - sum[p][l - 1]);
i = r + 1;
}
printf("%lld
", ans);
}
return 0;
}