正题

P1447

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题目大意

给出n,m，求$$2\times \sum _{i=1}^n\sum _{j=1}^m(gcd(i,j)-1)+n\times m$$

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解题思路

$$\begin{gathered} 2\times \sum _{i=1}^n\sum _{j=1}^m(gcd(i,j)-1)+n\times m \\ 2\times \sum _{i=1}^n\sum _{j=1}^{m}gcd(i,j)-n\times m \end{gathered}$$

那么求出${\sum }_{i=1}^n{\sum }_{j=1}^{m}gcd(i,j)$就好了

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^{m}gcd(i,j) \\ \sum _{d=1}^{n}d\sum _{i=1}^{n/d}\sum _{j=1}^{m/d}[gcd(i,j)=1] \\ \sum _{d=1}^{n}d\sum _{i=1}^{n/d}\sum _{j=1}^{m/d}\sum _{c|i,c|j}\mu (c) \\ \sum _{d=1}^{n}d\sum _{c=1}^{min(n/d,m/d)}\mu (c)\times \lfloor \frac{n}{cd}\rfloor \times \lfloor \frac{m}{cd}\rfloor \end{gathered}$$

考虑枚举cd，设k=cd

$$\sum _{k=1}^n\lfloor \frac{n}{k}\rfloor \lfloor \frac{m}{k}\rfloor \sum _{d|k}d\times \mu (k/d)$$

因为有$$\mu \times id=\phi$$

所以原式等于

$$\sum _{k=1}^n\lfloor \frac{n}{k}\rfloor \lfloor \frac{m}{k}\rfloor \phi (k)$$

可以整除分块

时间复杂度 $O(n+\sqrt{n})$

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code

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 100100
using namespace std;
ll n,m,w,ans,p[N],phi[N],prime[N];
const ll MX=1e5;
void work()
{phi[1]=1;for(ll i=2;i<=MX;++i){if(!p[i]){prime[++w]=i;phi[i]=i-1;}for(ll j=1;j<=w&&i*prime[j]<=MX;++j){p[i*prime[j]]=1;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else phi[i*prime[j]]=phi[i]*(prime[j]-1);}}for(ll i=2;i<=n;++i)phi[i]+=phi[i-1];return;
}
int main()
{scanf("%lld%lld",&n,&m);work();for(ll l=1,r=0;l<=min(n,m);l=r+1){r=min(n/(n/l),m/(m/l));ans+=(phi[r]-phi[l-1])*(n/l)*(m/l);}printf("%lld",ans*2-n*m);return 0;
}