正题

题目链接:https://www.luogu.com.cn/problem/P3312

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题目大意

定义$F(x)$表示$x$的约数和
给出$n,m,a$，求$$\sum _{i=1}^n\sum _{j=1}^m[F(gcd(i,j))\le a]\ast F(gcd(i,j))$$

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解题思路

首先我们不考虑$a$的限制，我们定义$g(i)={\sum }_{i|d}\mu (\frac{i}{d})\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{i}\rfloor$。那么我们有$$ans=\sum _{i=1}^{n}F(i)g(i)=\sum _{i=1}^{n}F(i)\sum _{i|d}\mu (\frac{i}{d})\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{i}\rfloor$$
$$\Rightarrow ans=\sum _{d=1}^n\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{i}\rfloor \sum _{i|d}F(i)\mu (\frac{d}{i})$$

显然我们可以计算${\sum }_{i|d}F(i)\mu (\frac{d}{i})$的前缀和来统计

考虑$a$的限制，我们发现对于$F(i)>a$那么有$F(i)=0$。我们可以将询问的$a$从小到大排序，然后按照$F(i)\mu (\frac{d}{i})$从小到大的顺序插入，因为要求前缀和，所以在树状数组里插入即可。

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define lowbit(x) (x&-x)
using namespace std;
const int N=1e5+10;
int T,n[N],m[N],a[N],p[N],t[N],nm;
int mu[N],f[N],g[N],pri[N],q[N],ans[N];
bool vis[N];
void prime(){int cnt=0;mu[1]=f[1]=1;for(int i=2;i<=nm;i++){if(!vis[i])vis[i]=1,pri[++cnt]=i,f[i]=g[i]=i+1,mu[i]=-1;for(int j=1;j<=cnt&&pri[j]*i<=nm;j++){vis[pri[j]*i]=1;if(i%pri[j]==0){mu[i*pri[j]]=0;g[i*pri[j]]=g[i]*pri[j]+1;f[i*pri[j]]=f[i]/g[i]*g[i*pri[j]];break;}mu[i*pri[j]]=-mu[i];f[i*pri[j]]=f[i]*f[pri[j]];g[i*pri[j]]=pri[j]+1;}}for(int i=1;i<=nm;i++)q[i]=i;
}
void Change(int x,int val){while(x<=nm){t[x]+=val;x+=lowbit(x);}return;
} 
int Ask(int x){int ans=0;while(x){ans+=t[x];x-=lowbit(x);}return ans;
}
bool cmp(int x,int y)
{return a[x]<a[y];}
bool cMp(int x,int y)
{return f[x]<f[y];}
int main()
{scanf("%d",&T);for(int i=1;i<=T;i++){scanf("%d%d%d",&n[i],&m[i],&a[i]);p[i]=i;}nm=100000;prime();sort(p+1,p+1+T,cmp);sort(q+1,q+1+nm,cMp);int jz=1;for(int i=1;i<=T;i++){int x=p[i];while(jz<=nm&&f[q[jz]]<=a[x]){for(int j=q[jz];j<=nm;j+=q[jz])Change(j,f[q[jz]]*mu[j/q[jz]]);jz++;}if(n[x]>m[x])swap(n[x],m[x]);for(int l=1,r;l<=n[x];l=r+1){r=min(n[x]/(n[x]/l),m[x]/(m[x]/l));ans[x]+=(n[x]/l)*(m[x]/l)*(Ask(r)-Ask(l-1));}}for(int i=1;i<=T;i++)printf("%d\n",ans[i]&(~(1<<31)));
}