LG3455[POI2007]ZAP-Queries【整除分塊+莫比烏斯反演】

前置知識：整除分塊

整除分塊

注：以下內容來自洛谷

題目大意

• 求$\sum_{i=1}^{a}\sum_{j=1}^{b}[gcd(x,y)=d]$
• 多組輸入
• $1\le d\le a,b\le 50000$

解題思路

數論函式

貼上程式碼：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int MX=50005,N=50505;
inline void Read(int &x){
    x=0;register char cc='\0';int fff=1;
    for(;cc<'0'||cc>'9';cc=getchar())if(cc=='-')fff=-1;
    for(;cc>='0'&&cc<='9';cc=getchar())x=(x<<1)+(x<<3)+(cc&15);
    x*=fff;
}
bool Pr[N];
int n,a,b,d,tot,p[N],mu[N];
void Prepare(){
    Pr[0]=Pr[1]=true;mu[1]=1;
    for(int i=2;i<=MX;++i){
        if(!Pr[i]){p[++tot]=i;mu[i]=-1;}
        for(int j=1;j<=tot&&i*p[j]<=MX;++j){
            Pr[i*p[j]]=true;
            if(i%p[j]==0)break;
            mu[i*p[j]]=mu[i]*mu[p[j]];
        }
    }
    for(int i=2;i<=MX;++i)
        mu[i]+=mu[i-1];
}
int main()
{
    freopen("LG3455.in","r",stdin);
    freopen("LG3455.out","w",stdout);
    Read(n);
    Prepare();
    while(n--){
        Read(a);Read(b);Read(d);
        a/=d;b/=d;if(a>b)swap(a,b);
        long long ans=0;
        for(int i=1,j=1;i<=a;i=j+1){
            j=min(a/(a/i),b/(b/i));
            ans+=1LL*(mu[j]-mu[i-1])*(a/i)*(b/i);
        } 
        printf("%lld\n",ans);       
    }
    return 0;
}