HDU 5382 GCD?LCM!

Description

First we define:
(1) $lcm(a,b)$, the least common multiple of two integers $a$ and $b$, is the smallest positive integer that is divisible by both $a$ and $b$. for example, $lcm(2,3)=6$ and $lcm(4,6)=12$.
(2) $gcd(a,b)$, the greatest common divisor of two integers $a$ and $b$, is the largest positive integer that divides both $a$ and $b$ without a remainder, $gcd(2,3)=1$ and $gcd(4,6)=2$.
(3) $[exp]$, $exp$ is a logical expression, if the result of $exp$ is $true$, then $[exp]=1$, else $[exp]=0$. for example, $[1+2\geq 3]=1$ and $[1+2\geq 4]=0$.

Now Stilwell wants to calculate such a problem:

$$F(n)=\sum_{i=1}^n\sum_{j=1}^n~[~lcm(i,j)+gcd(i,j)\geq n~] \\S(n)=\sum_{i=1}^nF(i)$$
Find $S(n)~mod~258280327$.

Solution

这题是非常有借鉴意义的，它揭露了数论里一个很重要的思想：把大问题化小再通过迭代搞没掉[一般用于这种大于等于的题目](记得ZJOI2016Day1 orzrzz再讲数论的时候某PE题就是用这样的思路，只是那时啥都不懂)

先带着把问题化小的思路进入题目。
观察到$n<=1,000,000$，说明可以线性递推

$$F(n)=\sum_{i=1}^n \sum_{j=1}^n [gcd(i,j)+lcm(i,j) \ge n]$$
$$=\sum_{i=1}^n \sum_{j=1}^n [gcd(i,j)+lcm(i,j) \ge n-1]\ - \ \sum_{i=1}^n \sum_{j=1}^n [gcd(i,j)+lcm(i,j) = n]$$
$$= F(n-1)+n*2-1+\sum_{i=1}^n \sum_{j=1}^n [gcd(i,j)+lcm(i,j) = n-1]$$

定义

$$T(n)=\sum_{i=1}^n \sum_{j=1}^n [gcd(i,j)+lcm(i,j) = n]$$
$$=>F(n)=F(n-1)+n*2-1-T(n-1)$$
化解$T(n)$：
$$T(n)=\sum_{i=1}^n \sum_{j=1}^n [gcd(i,j)+lcm(i,j) = n]$$
$$=\sum_{i=1}^n \sum_{j=1}^n[gcd(i,j)+i*j/gcd(i,j)=n]$$
$$=\sum_d \sum_{i=1}^{floor(n/d)} \sum_{j=1}^{floor(n/d)} [(i*j+1)d = n]$$
$$=\sum_d \sum_{i=1}^{floor(n/d)} \sum_{j=1}^{floor(n/d)} [gcd(i,j)=1][(i*j+1)*d=n]$$
$$=\sum_{d|n} \sum_{i=1}^{floor(n/d)} \sum_{j=1}^{floor(n/d)}[gcd(i,j)=1][i*j=n/d-1]................................(1)$$
定义$G(n)=\sum_{d|n}[gcd(d,n/d)=1]$
则$(1)$式$=G(n/d-1)$

$G(n)$是积性函数，推导如下：
当$n$的某因子次数为$\alpha_i$且$\alpha_i>1$，则$d$中要不包含$\alpha_i$个$p$，要不不包含$p$
所以对于$p$为素数，$(x,p)$，若$gcd(x,p)=1$，则$G(x*p)=G(x)*G(p)$
否则$G(x*p)=G(x)$

不如Eular筛一发得到$G$的值
然后枚举$d$，枚举$n/d$，可以$O(nln(n))$的得到$T$
然后由推导式线性得到$F$
然后前缀和得到
然后结束了

Code

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<time.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vec;
typedef priority_queue<int> pq;

#define pb push_back
#define ph push
#define fi first
#define se second

inline void Max(int &a,int b){if(a<b)a=b;}
inline void Min(int &a,int b){if(a>b||a==-1)a=b;}
template<class T>void rd(T &a){
    a=0;char c;
    while(c=getchar(),!isdigit(c));
    do a=a*10+(c^48);
        while(c=getchar(),isdigit(c));
}
template<class T>void nt(T x){
    if(!x)return;
    nt(x/10);
    putchar(48+x%10);
}
template<class T>void pt(T x){
    if(!x)putchar('0');
    else nt(x);
}
const int M=1e6+5;
const int P=258280327;
int prime[M],g[M],T[M],f[M],s[M],_,t;
bool mark[M];
inline void Mod_add(int &a,int b){if((a+=b)>=P)a-=P;}
inline void Eular(){
    g[1]=1;
    for(int i=2;i<M;++i){
        if(!mark[i])g[i]=2,prime[t++]=i;
        for(int j=0,p;j<t&&1ll*(p=prime[j])*i<M;++j){
            mark[i*p]=1;
            if(i%p==0){g[i*p]=g[i];break;}
            g[p*i]=g[i]<<1;
        }
    }
    for(int d=1;d<M;++d)
        for(int k=1;d*k<M;++k)
            Mod_add(T[d*k],g[k-1]);
    for(int i=1;i<M;++i)
        f[i]=((f[i-1]+(i<<1)-1-T[i-1])%P+P)%P;
    for(int i=1;i<M;++i)
        s[i]=(s[i-1]+f[i])%P;
}
inline void gao(){
    int n;cin>>n;
    pt(s[n]),putchar('\n');
}
int main(){
    for(cin>>_,Eular();_--;)gao();
}