lg3455

链接:lg3455

$ans=sum_{i}^a{sum_{j=1}^b{gcd(i,j)=d}}$

$ans=sum_{i}^{frac ad}{sum_{j=1}^{frac bd}{gcd(i,j)=1}}$

$ans=sum_{i}^{frac ad}{sum_{j=1}^{frac bd}{sum_{x|gcd(i,j)}^{}{mu(x)}}}$

$ans=sum_{x}^{min(frac ad,frac bd)}{mu(x) cdot sum_{i}^{frac ad}{sum_{j}^{frac bd}{x|gcd(i,j)}}}$

枚举ix jx

$ans=sum_(x)^{min(frac ad,frac bd)}{mu(x) cdot sum_{i}^{frac a{dx}}{sum_{j}^{frac b{dx}}}}$

$ans=sum_(x)^{min(frac ad,frac bd)}{mu(x) [frac a{dx}]cdot [frac b{dx}]}$

这样就可以O(n)做出来了,但是还是会T,这里用下除法分块就可以在O($sqrt n$)做出来


#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define rep(i,x,y) for(register int i=x;i<=y;++i)
#define repd(i,x,y) for(register int i=x;i>=y;--i)
#define ll long long
using namespace std;
const int N=1e6+7;
int mu[N],res[N],p[N],vis[N],cnt,a,b,d,T;
ll ans;
void (int n){
    mu[1]=1;
    rep(i,2,n){
        if(!vis[i])mu[i]=-1,p[++cnt]=i;
        for(int j=1;j<=cnt&&i<=n/p[j];++j){
            vis[i*p[j]]=1;
            if(i%p[j]==0)break;
            mu[i*p[j]]-=mu[i];
        }
    }
    rep(i,1,n)res[i]=res[i-1]+mu[i];
}
int main(){
	scanf("%d",&T);
	get_mu(500500);
	while(T--&&~scanf("%d%d%d",&a,&b,&d)){
		int l=1,r,len=min(a,b);ans=0;
		while(l<=len){
			r=min(a/(a/l),b/(b/l));
			ans+=1ll*(a/l/d)*1ll*(b/l/d)*1ll*(res[r]-res[l-1]);
			l=r+1;
		}
		printf("%lldn",ans);
	}
	return 0;
}

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