BZOJ2820: YY的GCD

莫比乌斯反演

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#define MAXN 10000000+10
#define INF 0x7f7f7f7f
#define LINF 0x7f7f7f7f7f7f7f7f
#define ll long long
#define pb push_back
#define ft first
#define sc second
#define mp make_pair
#define pil pair<int,ll>
#define pll pair<ll,ll>
using namespace std;
int f[MAXN],mu[MAXN],b[MAXN];
vector<int> p;
void getmu(){
    mu[1]=1;
    for(int i=2;i<MAXN;i++){
        if(!b[i]){
            p.pb(i);
            mu[i]=-1;
        }
        for(int j=0;j<p.size()&&p[j]*i<MAXN;j++){
            b[p[j]*i]=1;
            if(i%p[j]==0){
                mu[p[j]*i]=0;
                break;
            }
            mu[p[j]*i]=-mu[i];
        }
    }
    for(int i=0;i<p.size();i++){
        for(int j=1;j*p[i]<MAXN;j++){
            f[j*p[i]]+=mu[j];
        }
    }
    for(int i=1;i<MAXN;i++){
        f[i]+=f[i-1];
    }
}
int n,m;
void solve(){
    scanf("%d%d",&n,&m);
    if(n>m)swap(n,m);
    int lst;
    ll ans=0LL;
    for(int i=1;i<=n;i=lst+1){
        lst=min(n/(n/i),m/(m/i));
        ans+=1LL*(f[lst]-f[i-1])*(n/i)*(m/i);
    }
    printf("%lld
",ans);
}
int main()
{
    getmu();
    int T;
    scanf("%d",&T);
    while(T--){
        solve();
    }
    return 0;
}