对于给出的 (n) 个询问，每次求有多少个数对 ((x,y))，满足 (a le x le b)，(c le y le d)，且 (gcd(x,y) = k)，(gcd(x,y)) 函数为 (x) 和 (y) 的最大公约数。
前置知识
[sumlimits_{i=a}^{b} sumlimits_{j=c}^{d} [gcd(i,j)=k]
]
设
函数 (f(x,y)) 为：
[sumlimits_{i=1}^{x} sumlimits_{j=1}^{y} [gcd(i,j)=k]
]
那么，运用容斥原理，原式就 (= f(b,d) - f(a,d) - f(b,c) + f(a,c))
所以现在问题就转换为了求 (f) 函数。
[egin{aligned}
&sumlimits_{i=1}^{x} sumlimits_{j=1}^{y} [gcd(i,j)=k] \
=& sum_{i=1}^{lfloorfrac{n}{k}
floor}sum_{j=1}^{lfloorfrac{m}{k}
floor}[gcd(i,j)=1]\
=& sum_{i=1}^{lfloorfrac{n}{k}
floor}sum_{j=1}^{lfloorfrac{m}{k}
floor}varepsilon(gcd(i,j))\
=& sum_{i=1}^{lfloorfrac{n}{k}
floor}sum_{j=1}^{lfloorfrac{m}{k}
floor}sum_{d mid gcd(i,j) }mu(d)\
=& sum_{d=1 }mu(d)sum_{i=1}^{lfloorfrac{n}{k}
floor}[d mid i]sum_{j=1}^{lfloorfrac{m}{k}
floor}[d mid j]\
=&sum_{d=1}mu(d)lfloorfrac{n}{kd}
floorlfloorfrac{m}{kd}
floor
end{aligned}
]
我们可以使用数论分块进行求解
代码：
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<typename T>inline void read(T &FF){
T RR=1;FF=0;char CH=getchar();
for(;!isdigit(CH);CH=getchar())if(CH=='-')RR=-1;
for(;isdigit(CH);CH=getchar())FF=(FF<<1)+(FF<<3)+(CH^48);
FF*=RR;
}
const int N=50010;
int prim[N],mu[N],sum[N],cnt,k,T;
bool vis[N];
void init(){
mu[1]=1;
for(register int i=2;i<N;i++){
if(!vis[i]){
mu[i]=-1;
prim[++cnt]=i;
}
for(register int j=1;j<=cnt&&i*prim[j]<N;j++){
vis[i*prim[j]]=1;
if(i%prim[j]==0)break;
mu[i*prim[j]]=-mu[i];
}
}
for(register int i=1;i<N;i++)sum[i]=sum[i-1]+mu[i];
}//莫比乌斯反演的板子
ll calc(int a,int b){
ll ans=0;
for(register int l=1,r;l<=min(a,b);l=r+1){
r=min(a/(a/l),b/(b/l));
ans+=(1ll*a/l)*(1ll*b/l)*(sum[r]-sum[l-1]);
}
return ans;
}
int main(){
init();
for(read(T);T--;){
int a,b,c,d;
read(a);read(b);read(c);read(d);read(k);
a--;c--;a/=k;b/=k;c/=k;d/=k;
printf("%lld
",calc(b,d)-calc(b,c)-calc(a,d)+calc(a,c));
}
return 0;
}