[NOI2010]能量采集

题链

ans=2*（Σ(1<=i<=n)Σ(1<=j<=m)GCD(i,j)）-m*n，

经典的反演。

#include<bits/stdc++.h>
#define N 100007
#define LL long long
LL ans,sum[N],p[N>>2],u[N],tot;
using namespace std;
void get() {
    sum[1]=1;
    for (int i=2;i<N;i++) {
        if(!u[i]) p[++tot]=i,u[i]=i-1;
        for (int j=1;j<=tot&&i*p[j]<N;j++) 
            if (i%p[j]) u[i*p[j]]=u[i]*(p[j]-1);
            else {u[i*p[j]]=u[i]*p[j];break;}
        sum[i]=sum[i-1]+u[i];
    }
}
int n,m,k;
int main () {
    scanf("%d%d",&n,&m);
    get();
    k=min(n,m);
    for (int i=1,last;i<=k;i=last+1){
        last=min(n/(n/i),m/(m/i));
        ans+=(sum[last]-sum[i-1])*(n/i)*(m/i);
    }
    printf("%lld\n",ans*2-(LL)n*m);
}