[bzoj2154]Crash的数字表格(mobius反演)

题意：$sumlimits_{i = 1}^n {sumlimits_{j = 1}^m {lcm(i,j)} } $

解题关键：

$sumlimits_{i = 1}^n {sumlimits_{j = 1}^m {lcm(i,j)} }  = sumlimits_{i = 1}^n {sumlimits_{j = 1}^m {frac{{i*j}}{{gcd (i,j)}}} } $

枚举gcd，上式化为：

$sumlimits_{d = 1}^{min (n,m)} {dsumlimits_{egin{array}{*{20}{c}}
{gcd (i,j) = = 1}\
{1 < = i < = n/d}\
{1 < = j < = m/d}
end{array}} {i*j} } $

令

$f(n,m,k) = sumlimits_{egin{array}{*{20}{c}}
{gcd (i,j) = = k}\
{1 < = i < = n}\
{1 < = j < = m}
end{array}} {i*j} $

由于

$sum(n,m) = sumlimits_{egin{array}{*{20}{c}}
{1 < = i < = n}\
{1 < = j < = m}
end{array}} {i*j} = frac{{n(n + 1)}}{2}frac{{m(m + 1)}}{2}$

$egin{array}{l}
F(n,m,k) = sumlimits_{k|d} {f(n,m,d) = sumlimits_{egin{array}{*{20}{c}}
{1 < = i < = n}\
{1 < = j < = m}\
{k|gcd (i,j)}
end{array}} {i*j} = {k^2}sum(leftlfloor {frac{n}{k}} ight floor ,leftlfloor {frac{m}{k}} ight floor )} \
f(n,m,k) = sumlimits_{k|d} {u(frac{d}{k})F(n,m,d)}
end{array}$

而此题中，$k==1$,

则，

$egin{array}{l}
f(n,m,1) = sumlimits_{d = 1}^{min (n,m)} {u(d)F(n,m,d)} \
= sumlimits_{d = 1}^{min (n,m)} {u(d){d^2}sum(leftlfloor {frac{n}{d}} ight floor ,leftlfloor {frac{m}{d}} ight floor )} \
ans = sumlimits_{d = 1}^{min (n,m)} {d*f(leftlfloor {frac{n}{d}} ight floor ,leftlfloor {frac{m}{d}} ight floor ,1)}
end{array}$

 求解ans和$f$函数复杂度都是$O(sqrt n )$，所以最终复杂度为$O(n)$。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<iostream>
#define Sum(x,y) (x*(x+1)/2%mod*(y*(y+1)/2%mod)%mod)
using namespace std;
typedef long long ll;
const ll mod=20101009;
const ll maxn=10000000+200;
ll n,m;
bool vis[maxn];
ll prime[maxn],mu[maxn],sum1[maxn];
void init_mu(ll n){
    ll cnt=0;
    mu[1]=1;
    for(ll i=2;i<n;i++){
        if(!vis[i]){
            prime[cnt++]=i;
            mu[i]=-1;
        }
        for(ll j=0;j<cnt&&i*prime[j]<n;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){mu[i*prime[j]]=0;break;}
            else { mu[i*prime[j]]=-mu[i];}
        }
    }
    sum1[0]=0;
    for(ll i=1;i<n;i++) sum1[i]=(sum1[i-1]+1ll*mu[i]*i*i)%mod;
}
ll calf(ll n,ll m){
    ll ans=0,pos,len=min(n,m);
    for(ll i=1;i<=len;i=pos+1){
        pos=min(n/(n/i),m/(m/i));
        //ans+=(sum1[pos]-sum1[i-1])%mod*((n/i)*((n/i)+1)/2%mod*(((m/i)+1)*(m/i)/2%mod)%mod);
        ans+=(sum1[pos]-sum1[i-1])%mod*Sum(n/i,m/i)%mod;//最好用函数写出 
        ans%=mod;
        
    }
    return ans;
}
ll calres(ll n,ll m){
    ll ans=0,pos,len=min(n,m);
    for(ll i=1;i<=len;i=pos+1){
        pos=min(n/(n/i),m/(m/i));
        ans+=(i+pos)*(pos-i+1)/2%mod*calf(n/i,m/i)%mod;
        ans%=mod;
    }
    return (ans%mod+mod)%mod;
}

int main(){
    scanf("%lld%lld",&n,&m);
    init_mu(min(n,m)+10);
    printf("%lld
",calres(n,m));
    return 0;
}