【LuoguP1829】Crash的数字表格

题目链接

题目描述

求：

$$\sum_{i=1}^n\sum_{j=1}^mlcm(i,j)$$
$lcm(i,j)$ 表示i，j的最小公倍数

题解

先是一步简单的变形:

$$\sum_{i=1}^n\sum_{j=1}^m\dfrac {ij}{gcd(i,j)}$$
常用的套路就是枚举gcd(i,j):
$$\sum_{d=1}^{min(n,m)}d\sum_{i=1}^{\frac nd}\sum_{j=1}^{\frac md}[gcd(i,j)=1](ij)$$
然后其实有两种做法:
1.
$$\sum_{d=1}^{min(n,m)}d\sum_{i=1}^{\frac nd}\sum_{j=1}^{\frac md}(ij)\sum_{d'|gcd(i,j)}\mu(d')$$
设 $F(x)表示\sum_{i=1}^xi$
$$\sum_{d=1}^{min(n,m)}d\sum_{d'=1}^{min(\frac nd,\frac md)}\mu(d')*d'^2*F(\frac n{dd'})*F(\frac m{dd'})$$
设dd’=T
$$\sum_{T=1}^{min(n,m)}F(\frac nT)*F(\frac mT)*T\sum_{d'|T}d'\mu(d')$$
后面那个玩意很容易筛，前面的也预处理一下，再用下求和公式就可以了。
$O(\sqrt n)$
2.
我们直接设 $f(x,y,k)=\sum_{i=1}^x\sum_{j=1}^y[gcd(i,j)=k]ij$
由定义可设 $F(x,y,k)=\sum_{i=1}^x\sum_{j=1}^y[gcd(i,j)|k]ij$
(这里函数并不是说状态量多了，前两个只是范围限制，k才是真正变量)
然后原式化为:
$$\sum_{d=1}^{min(n,m)}d*f(\frac nd,\frac md,1)$$
$$\sum_{d=1}^{min(n,m)}d*\sum_{1|t}\mu(\frac t1)F(\frac nd,\frac md,t)$$
$$\sum_{d=1}^{min(n,m)}d*\sum_{t=1}^{min(\frac nd,\frac md)}\mu(t)F(\frac nd,\frac md,t)$$
$$\sum_{d=1}^{min(n,m)}d*\sum_{t=1}^{min(\frac nd,\frac md)}\mu(t)*t^2\dfrac {(n/d)∗(n/d+1)}2*\dfrac {(m/d)∗(m/d+1)}2$$
$$\sum_{d=1}^{min(n,m)}d*\dfrac {(n/d)∗(n/d+1)}2*\dfrac {(m/d)∗(m/d+1)}2\sum_{t=1}^{min(\frac nd,\frac md)}\mu(t)*t^2$$

后面那坨预处理，分块搞就可以了
复杂度$O(\sqrt n^2)=O(n)$,但预处理常数较小，所以跑得比法1快……

法1的代码:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int mod=20101009;
const int N=1e7+10;
int pri[N];bool vis[N];int mu[N];
int sum[N];int cnt=0;
int f[N];
typedef long long ll;
inline void prepare()
{
    mu[1]=1;f[1]=1;
    vis[1]=1;
    for(register int i=2;i<N;i++)
    {
        if(!vis[i]) {pri[++cnt]=i;mu[i]=-1;f[i]=1-i+mod;}
        for(register int j=1;j<=cnt&&(1ll*i*pri[j]<N);++j)
        {
            register int x=i*pri[j];
            vis[x]=1;
            if(i%pri[j]==0) {
                mu[i]=0;f[x]=f[i];
                break;
            }
            mu[x]=-mu[i];
            f[x]=(1ll*f[i]*f[pri[j]])%mod;
        }
    }
    sum[1]=1;
    for(register int i=1;i<N;++i) {f[i]=f[i-1]+1ll*f[i]*i%mod;if(f[i]>=mod) f[i]-=mod;}// T*sigma(d*mu[d]) 
    for(register int i=1;i<N;++i) {sum[i]=sum[i-1]+i;if(sum[i]>=mod) sum[i]-=mod;}
}
int main()
{
    int n,m;prepare();
    scanf("%d %d",&n,&m);register int l,r;
    if(n>m) swap(n,m);register int ans=0;
    for(l=1;l<=n;l=r+1){
        r=min(n/(n/l),m/(m/l));
        register int S=f[r]-f[l-1]+mod;if(S>=mod) S-=mod;
        ans+=1ll*S*sum[n/l]%mod*sum[m/l]%mod;
        if(ans>=mod) ans-=mod;
    }
    printf("%d\n",ans);
}

法2的代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
typedef long long ll;
const int mod=20101009;
const int N=1e7+100;
int mu[N];ll sum[N];
int pri[N];bool vis[N];int cnt=0;
inline void prepare()
{
    mu[1]=1;vis[1]=1;
    for(register int i=2;i<N;i++)
    {
        if(!vis[i]) {pri[++cnt]=i;mu[i]=-1;}
        for(register int j=1;j<=cnt&&(1ll*i*pri[j]<N);j++)
        {
            register int x=i*pri[j];
            vis[x]=1;
            if(i%pri[j]==0) {mu[x]=0;break;}
            mu[x]=-mu[i];
        }
    }
    for(register ll i=1;i<N;i++) sum[i]=(1ll*mu[i]*(i*i%mod)+sum[i-1]+mod)%mod;
}
inline ll Get(int n,int m)
{
    return (((1ll*n*(n+1)/2)%mod)*((1ll*m*(m+1)/2)%mod))%mod;
}
inline ll calc(int n,int m)
{
    register int l,r;
    register ll res=0;
    for(l=1;l<=n;l=r+1)
    {
        r=min(n/(n/l),m/(m/l));
        register ll S=sum[r]-sum[l-1];if(S<0) S+=mod;
        res+=(1ll*S*Get(n/l,m/l))%mod;
        if(res>=mod) res-=mod;
    }
    return res;
}
int main()
{
    int n,m;prepare();
    scanf("%d %d",&n,&m);
    if(n>m) swap(n,m);
    register ll l,r;
    register ll ans=0;
    for(l=1;l<=n;l=r+1)
    {
        r=min(n/(n/l),m/(m/l));
        ans+=(((l+r)*(r-l+1)/2)%mod*calc(n/l,m/l))%mod;
        if(ans>=mod) ans-=mod;
    }
    printf("%lld\n",ans);
}