[POI2007] ZAP-Queries, Luogu P3455, Mobius Inversion

The subject

      This question is very obvious, directly use Mobius inversion to find gcd(x,y)=d.

      You can set a function f(d) to represent the logarithm of gcd of d in 1~a and 1~b.

      Then to learn a routine is to set a summation function F(d) to represent the logarithm of gcd of a multiple of d in 1~a and 1~b.

      Obviously F(d)=f(d)+f(2d)+f(3d)...

      Of course, F(d) can also be calculated by a formula =(a/d)*(b/d).

      The next step is to see how the division is optimized in blocks, and specifically look at the code because I don't know how to express it in mathematical symbols. . emm

Code <don't mind if there are not many functions>

#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int n;
int d,a,b;
int p[50010];
int miu[50010];
bool tf[50010];
int sum[50010];

long long solve(){
	long long ans=0;
	int l = 1, r;
	a/=d;b/=d;//First divide a d, the problem is transformed into how many pairs of coprime logarithms are there from 1 to a and 1 to b
	while(l<=min(a,b)){//l enumerates the current accumulation to f(lx)
		r=min(a/(a/l),b/(b/l));//The division is divided into blocks to calculate the right endpoint, at this time you can know the F(ix) between l~r (l<=i< =r) is the same value.
		ans=ans+(long long)(a/l)*(b/l)*(sum[r]-sum[l-1]);//So multiply the sum of miu values ​​of this paragraph, you can use the prefix sum to calculate
		l=r+1;
	}
	return ans;
}

int main(){
	scanf("%d",&n);
	miu[1]=1;
	sum[1]=1;
	for(int i=2;i<=50000;i++){//Linear sieve Möbius function
		if(!tf[i]) {miu[i]=-1;p[++p[0]]=i;}
		for(int j=1;j<=p[0] && (long long)i*p[j]<=50000;j++){
			tf[p[j]*i]=true;
			if(i%p[j]==0) {miu[i*p[j]]=0;break;}
			miu[i*p[j]]=-miu[i];
		}
		sum[i]=sum[i-1]+miu[i];//The prefix sum is calculated to facilitate the subsequent solution
	}
	while(n--){
		scanf("%d %d %d",&a,&b,&d);
		printf("%lld\n",solve());//Enter the function to solve
	}
}