lg3455

链接:lg3455

题解

  • $ans=\sum_{i}^a{\sum_{j=1}^b{gcd(i,j)=d}}$
  • $ans=\sum_{i}^{\frac ad}{\sum_{j=1}^{\frac bd}{gcd(i,j)=1}}$
  • $ans=\sum_{i}^{\frac ad}{\sum_{j=1}^{\frac bd}{\sum_{x|gcd(i,j)}^{}{\mu(x)}}}$
  • $ans=\sum_{x}^{min(\frac ad,\frac bd)}{\mu(x) \cdot \sum_{i}^{\frac ad}{\sum_{j}^{\frac bd}{x|gcd(i,j)}}}$
  • 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
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#include<cstdio>
#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 get_mu(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("%lld\n",ans);
}
return 0;
}