洛谷 P2568 GCD

题目描述: 给定整数N，求1<=x,y<=N且Gcd(x,y)为素数的数对(x,y)有多少对.

思路:莫比乌斯反演的套路题，先设f(n)表示为gcd(i,j)=n的对数，g(n)表示为n | gcd(i,j)的对数,则g(n) = ∑_n|df(d)=n/d*n/d

则f(d) = ∑_t=1g(t*d)*mu(t)  (mu代表莫比乌斯函数,1<=t<=n/d)， 然后只要枚举小于等于n的素数就行了

#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=1e7+5;
typedef long long ll;
int dir[4][2]={-1,0,1,0,0,-1,0,1};
int mu[maxn];
int sum[maxn];
int prime[maxn];
int tot;
int vis[maxn];
void table(){
    memset(mu,0,sizeof(mu));
    memset(sum,0,sizeof(sum));
    memset(vis,0,sizeof(vis));
    tot=0;
    mu[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&&prime[j]*i<maxn;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0) break;
            else mu[i*prime[j]]=-mu[i];
        }
    }
}
int main(){
    ll n;
    ios::sync_with_stdio(false);
    table();
    while(cin>>n){
        ll res=0;
        for(ll i=0;i<tot&&prime[i]<=n;i++){
            ll k=prime[i];
            ll m=n/k;
            ll ans=0;
            for(ll x=1,y;x<=m;){
                y=m/(m/x);
                ans+=(m/x)*(m/x)*(sum[y]-sum[x-1]);
                x=y+1;
            }
            res+=ans;
        }
        cout<<res<<endl;
    }
}