51nod1220约数之和

51nod1220约数之和

Time Limit 3s Memory Limit 128KB
Description
$\sigma(x)$表示x的约束和
求$\sum_{i=1}^n \sum _{j=1}^n \sigma(ij)$
Input
一个数n(0$\lt$n$\leq 10^9$)
Output
一个数答，对$10^9+7$取模。
Sample Input

1000

Sample Output

576513341

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解题思路

莫比乌斯反演，杜教筛的各种技能
定义$[x]$的取值为1(x为真)否则为0

$$Ans=\sum_{i=1}^n \sum_{j=1}^n \sum_{d|ij}d \\$$
$d|ij$能不能化为一个$d'|j$呢，想到$d|ij$等同于$\frac d{\gcd(i,d)}|j$

d|ij要满足$\exists_{xy=d}[x|i][y|j]$，x取$\gcd(i,d)$能供给完$[x|i]$也就是说，这时$\gcd(y,i)=1$当且仅当$y|j$时$d|ij$成立，即$d|ij \leftrightarrow \frac d{\gcd(i,d)}|j$

$\therefore$

$$\begin{align} Ans & =\sum_{i=1}^n \sum_{j=1}^n \sum_{d=1}^{ij}d[\frac d{\gcd(i,d)}|j] \\ & =\sum_{i=1}^n \sum_{d=1}^{in}d\lfloor\frac {n\gcd(i,d)}d\rfloor \\ & =\sum_{d'=1}^n\sum_{i=1}^n \sum_{d=1}^{in}[\gcd(i,d)=d']d\lfloor\frac {nd'}d\rfloor \ \ \ \ \ \ (d'表示gcd(i,d)) \end{align}$$

$\gcd(i,d)=d'$?
考虑化为$\gcd(,)=1$很简单啊，用$\frac i{d'}*d'$代替i，用$\frac d{d'}*d'$代替d

$$Ans=\sum_{d'=1}^n\sum_{\frac i{d'}=1}^{\lfloor\frac n{d'}\rfloor} \sum_{\frac d{d'}=1}^{\frac i{d'}n}[\gcd(\frac i{d'},\frac d{d'})=1]\frac d{d'}*d'\lfloor\frac {d'}d*n\rfloor$$
用i代替$\frac i{d'}$，用d代替$\frac d{d'}$
$$\begin{align} Ans & =\sum_{d'=1}^n\sum_{i=1}^{\lfloor\frac n{d'}\rfloor} \sum_{d=1}^{in}[\gcd(i,d)=1] dd'\lfloor\frac nd\rfloor \\ & =\sum_{d'=1}^nd'\sum_{i=1}^{\lfloor\frac n{d'}\rfloor} \sum_{d=1}^n[\gcd(i,d)=1] d\lfloor\frac nd\rfloor \\ & =\sum_{i=1}^n\sum_{d'=1}^{\lfloor\frac ni\rfloor}d' \sum_{d=1}^n[\gcd(i,d)=1] d\lfloor\frac nd\rfloor \\ & =\sum_{i=1}^n\frac 12\lfloor\frac ni\rfloor(\lfloor\frac ni\rfloor+1) \sum_{d=1}^n[\gcd(i,d)=1] d\lfloor\frac nd\rfloor \\ \end{align}$$
$\gcd(i,d)=1$怎么处理呢，想到$\sum_{d|n}\mu(d)=[n=1]$那么
$$[\gcd(i,d)=1]=\sum_{k|\gcd(i,d)}\mu(k)=\sum_k [k|i][k|d]\mu(k)$$
$\therefore$
$$\begin{align} Ans & =\sum_{i=1}^n\frac 12\lfloor\frac ni\rfloor(\lfloor\frac ni\rfloor+1) \sum_{d=1}^n d\lfloor\frac nd\rfloor\sum_{d'=1}^n [d'|i][d'|d]\mu(d') \\ & =\sum_{d'=1}^n\mu(d')\sum_{\frac i{d'}=1}^{\lfloor\frac n{d'}\rfloor}\frac i{d'}*d'\lfloor\frac n{\frac i{d'}*d'}\rfloor\sum_{\frac d{d'}=1}^{\lfloor\frac n{d'}\rfloor}\frac 12\lfloor\frac n{\frac d{d'}*d'}\rfloor(\lfloor\frac n{\frac d{d'}*d'}\rfloor+1) \\ & =\sum_{d'=1}^nd'\mu(d')\sum_{i=1}^{\lfloor\frac n{d'}\rfloor} i\lfloor \frac n{id'}\rfloor\sum_{d=1}^{\lfloor\frac n{d'}\rfloor}\frac 12\lfloor\frac n{dd'}\rfloor(\lfloor\frac n{dd'}\rfloor+1) \end{align}$$
设
$$f(x)=x\mu(x)$$
$$g(x)=\sum_{i=1}^xi\lfloor\frac xi\rfloor$$
$$h(x)=\sum_{i=1}^x\frac 12\lfloor\frac xi\rfloor(\lfloor\frac xi\rfloor+1)$$

其实h和g是等价的

$$\begin{align} \sum_{i=1}^x\frac 12\lfloor\frac xi\rfloor(\lfloor\frac xi\rfloor+1) & =\sum_{i=1}^x\sum_{j=1}^{\lfloor\frac xi\rfloor}j(i的出现次数和j的出现次数是一样的)\\ & =\sum_{i=1}^x\sum_{j=1}^{\lfloor\frac xi\rfloor}i\\ & =\sum_{i=1}^x\lfloor\frac xi\rfloor i \end{align}$$

$\therefore$

$$\begin{align} Ans & =\sum_{d=1}^n f(d)\cdot g(\lfloor\frac nd\rfloor)h(\lfloor\frac nd\rfloor) \\ & =\sum_{d=1}^n g(\lfloor\frac nd\rfloor)^2f(d) \end{align}$$

我们想，$\lfloor\frac nd\rfloor$只有$2\sqrt n$种

当$i<=\sqrt n$时，$i\in[1,\sqrt n]$
当$i>\sqrt n$时,$\frac ni<\sqrt n$
$\therefore \lfloor\frac ni\rfloor$只有$2\sqrt n$种
对于$i$，有$j=max\{x|\lfloor\frac nx\rfloor=\lfloor\frac ni\rfloor\}=\lfloor\frac n{\lfloor \frac ni\rfloor}\rfloor$

所以，分块处理
$对于i，有\forall_{j\in[\lfloor\frac n{\lfloor \frac n{i-1}\rfloor}\rfloor+1,\lfloor\frac n{\lfloor \frac ni\rfloor}\rfloor]}\lfloor\frac nj\rfloor=i$
只需维护$s(x)=\sum_{i=1}^xf(x)$即可，$g(\lfloor \frac ni\rfloor)$直接算
有$O(n)=\sum_{\lfloor\frac ni\rfloor}O(\sqrt \frac ni)=O(n^{\frac 34})$

对于S(x)
$\sum_{d|n}\mu(d)=[n=1]$
那么

$$\begin{align} 1 & = \sum_{i=1}^ni\sum_{d|i}\mu(d)\\ & = \sum_{d=1}^n\sum_{i=1}^{\lfloor\frac ni\rfloor}id\mu(d)\\ & = \sum_{d=1}^n\sum_{i=1}^{\lfloor\frac ni\rfloor}di\mu(i)\\ & = \sum_{d=1}^nd\sum_{i=1}^{\lfloor\frac ni\rfloor}f(i)\\ & = \sum_{d=1}^nds(\lfloor\frac nd\rfloor) \end{align}$$
$$\therefore s(n)=1-\sum_{d=2}^n ds(\lfloor\frac nd\rfloor)(将d=1项移项)$$
同样，分块处理$\because\lfloor\frac n4\rfloor=\lfloor\frac{\lfloor\frac n2\rfloor}2\rfloor$，所以直接搜会有很多重复的项，哈希判重，$O(n)=\sum_{\lfloor\frac ni\rfloor}O(\sqrt \frac ni)=o(n^{\frac 34})$
预处理前$n^{\frac 23}$项，复杂度可降到$O(n^{\frac 23})$

总复杂度$O(n^{\frac 34})$

#include<cstring>
#include<cstdio>
#include<cctype>
#define N 3001001
#define lim 43007
#define mo 1000000007
#define ny 500000004
using namespace std;
typedef long long ll;
ll n,s[N],sum[lim],hash[lim],mu[N],pr[N];
bool bz[N];
int get(ll x){
    int y=(int)(x%lim);
    while(hash[y] && hash[y]!=x)y++,y=y==lim?0:y;return y;
}
ll S(ll a){
    if(a<N)return s[a];
    int x=get(a);
    if(hash[x])return sum[x];
    hash[x]=a;ll ans=1;
    for(ll i=2,j;i<=a;i=j+1){
        j=a/(a/i);
        ans=(ans-(j+i)%mo*((j-i+1)%mo)%mo*ny%mo*S(a/i)%mo+mo)%mo;
    }sum[x]=ans;return ans;
}
ll g(ll a){
    ll ans=0;
    for(ll i=1,j;i<=a;i=j+1){
        j=a/(a/i);
        ans=(ans+a/i%mo*((i+j)%mo*((j-i+1)%mo)%mo*ny%mo))%mo;
    }return ans;
}
int main(){
    s[1]=mu[1]=1;
    for(ll i=2;i<N;i++){
        if(!bz[i])mu[pr[++pr[0]]=i]=-1;
        s[i]=(s[i-1]+mu[i]*i%mo+mo)%mo;
        for(int j=1;j<=pr[0] && i*pr[j]<N;j++){
            bz[i*pr[j]]=1;mu[i*pr[j]]=-mu[i];
            if(i%pr[j]==0){mu[i*pr[j]]=0;break;}
        }
    }
    scanf("%lld",&n);
    ll ans=0;
    for(ll i=1,j,las=0,now,t;i<=n;i=j+1,las=now){
        j=n/(n/i);now=S(j);t=g(n/i);
        ans=(ans+(now-las+mo)%mo*t%mo*t%mo)%mo;
    }
    printf("%lld",ans);
}