洛谷 P3455 [POI2007]ZAP-Queries (莫比乌斯反演)

题意: 给定a，b，d求gcd(x,y)=d的对数(1<=x<=a,1<=y<=b)

思路：按照套路来先设f(n)为gcd(x,y)=n的对数,g(n)表示为 n | gcd(x,y)的对数,则g(n)=∑_n|df(d)=a/n*b/n

f(n)=∑_n|dg(d)*mu(d/n),令t=d/n则f(n)=∑_t=1g(t*n)*mu(t)，然后求f(d)就行了

#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<vector>
#include<cstdio>
#include<queue>
#include<map>
#include<set>
#include<math.h>
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=1e5+5;
typedef long long ll;
int dir[4][2]={-1,0,1,0,0,-1,0,1};
int prime[maxn/10];
int tot;
int mu[maxn];
int sum[maxn];
int vis[maxn];
void table(){
    memset(vis,0,sizeof(vis));
    memset(sum,0,sizeof(sum));
    memset(mu,0,sizeof(mu));
    tot=0;
    mu[1]=1;
    vis[0]=vis[1]=1;
    sum[1]=1;
    for(int i=2;i<maxn;i++){
        if(!vis[i]){
            prime[tot++]=i;
            mu[i]=-1;
        }
        sum[i]=sum[i-1]+mu[i];
        for(int j=0;j<tot&&i*prime[j]<maxn;j++){
             vis[i*prime[j]]=1;
             if(i%prime[j]){
                 mu[i*prime[j]]=-mu[i];
             }else break;
        }
    }
}
int main(){
    ll t;
    ios::sync_with_stdio(false);
    table();
    cin>>t;
    while(t--){
       ll a,b,d;
       cin>>a>>b>>d;
       a/=d;b/=d;
       ll limit=min(a,b);
       ll ans=0;
       for(ll x=1,y;x<=limit;){
               //cout<<x<<endl;
              y=min(a/(a/x),b/(b/x));
              ans+=(a/x)*(b/x)*(sum[y]-sum[x-1]);
              x=y+1;
       }
       cout<<ans<<endl;
    }
}