題目

給定$n,m$，求$1\leq x\leq n$,$1\leq y\leq m$滿足$gcd(x,y)=d$的$(x,y)$個數

分析

設$f(d)=\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=d]$ $F(d)=\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor$

程式碼

    #include <cstdio>
    #include <vector>
    #define rr register
    #define N 50001
    using namespace std;
    int sum[N],v[N],mobius[N];
    inline
 
 int in(){
    	rr int ans=0; rr char c=getchar();
    	while (c<48||c>57) c=getchar();
    	while (c>47&&c<58) ans=(ans<<3)+(ans<<1)+c-48,c=getchar();
    	return ans;
    }
    void print(int ans){
    	if (ans>9) print(ans/10);
    	putchar(ans%10+48);
    }
    void
 
 prepare(int n){
    	mobius[1]=1; vector<int>prime;
    	for (rr int i=2;i<=n;++i){
    		if (!v[i]) v[i]=i,mobius[i]=-1,prime.push_back(i);
    		for (rr int j=0;j<prime.size();++j){
    			if (prime[j]>v[i]||prime[j]>n/i) break;
    			v[i*prime[j]]=prime[j];
    			if (i%prime[j]==0) break;
    			mobius[i*prime[j]]=-mobius[i];
    		}
    	}
    	for (rr int i=1;i<=n;i++) sum[i]=sum[i-1]+mobius[i];//線性篩求莫比烏斯字首和
    }
    int main(){
    	rr int t=in(); prepare(N-1);
    	while (t--){
    		rr int n=in(),m=in(),d=in();
    		if (n>m) n^=m,m^=n,n^=m;
    		rr long long ans=0;
    		for (rr int l=1,r;l<=n;l=r+1){//整除分塊
    			r=min(n/(n/l),m/(m/l));
    			ans+=(n/l/d)*(m/l/d)*(sum[r]-sum[l-1]);//計算答案字首和
    		}
    		print(ans); putchar(10);
    	}
    	return 0;
    }