正题

题目链接:https://www.luogu.com.cn/problem/P1829

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题目大意

给出$n,m$求$$\sum _{i=1}^n\sum _{j=1}^{m}lcm(i,j)$$

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解题思路

$$\sum _{i=1}^n\sum _{j=1}^m\frac{ij}{gcd(i,j)}$$
$$\sum _{x=1}^{n}x\times (\sum _{i=1}^{\lfloor \frac{n}{x}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{x}\rfloor }ij\times [gcd(i,j)==1])$$
$$\sum _{x=1}^{n}x\times \sum _{x|d}\mu (\frac{d}{x})d^2\ast \sum _{i=1}^{\lfloor \frac{n}{x}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{x}\rfloor }ij$$
单独拎出后面的来看
$$\sum _{x|d}\mu (\frac{d}{x})d^2\ast \sum _{i=1}^{\lfloor \frac{n}{x}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{x}\rfloor }ij$$
$$\sum _{x|d}\mu (\frac{d}{x})d^2\ast \sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }i\ast \sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }j$$
然后整除分块处理这一个
之后再在外面套一个整除分块即可。

时间复杂度$O(n+m)$

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll N=1e7+10,XJQ=20101009;
ll n,m,mu[N],pri[N],g[N],cnt,ans;
bool v[N];
void ycl(ll n){mu[1]=1;for(ll i=2;i<=n;i++){if(!v[i])pri[++cnt]=i,mu[i]=-1;for(ll j=1;j<=cnt&&i*pri[j]<=n;j++){v[i*pri[j]]=1;if(i%pri[j]==0)break;mu[i*pri[j]]=-mu[i];}}for(ll i=1;i<=n;i++)mu[i]=(mu[i-1]+mu[i]*i*i%XJQ)%XJQ;return;
}
ll solve(ll n,ll m){ll ans=0;for(ll l=1,r;l<=n;l=r+1){r=min(n/(n/l),m/(m/l));ll q=n/l,p=m/l;ans=(ans+(q*(q+1)/2%XJQ)*(p*(p+1)/2%XJQ)%XJQ*(mu[r]-mu[l-1]+XJQ)%XJQ)%XJQ;}return ans;
}
int main()
{scanf("%lld%lld",&n,&m);if(n>m)swap(n,m);ycl(m);for(ll l=1,r;l<=n;l=r+1){r=min(n/(n/l),m/(m/l));(ans+=(r+l)*(r-l+1)/2%XJQ*solve(n/l,m/l)%XJQ)%=XJQ;}printf("%lld\n",ans);
}