[Codeforces235E]Number Challenge（莫比乌斯反演）_codeforces 235e number challenge-优快云博客

本文链接：https://blog.youkuaiyun.com/Clove_unique/article/details/64157271

题目描述

题解

看到这道题有没有想到sdoi的约数个数和？
没错真的是类似的

首先考虑 $d(a*b*c)$ 是多少
有一个结论：

d (a * b * c) = \sum i | a \sum j | b \sum k | c [(i, j) = 1] [(j, k) = 1] [(i, k) = 1]

$d(a*b*c)=\sum\limits_{i|a}\sum\limits_{j|b}\sum\limits_{k|c}[(i,j)=1][(j,k)=1][(i,k)=1]$
然后将这个式子带入

\sum i = 1 a \sum j = 1 b \sum k = 1 c \sum x | a \sum y | b \sum z | c [(x, y) = 1] [(y, z) = 1] [(x, z) = 1]

$\sum\limits_{i=1}^a\sum\limits_{j=1}^b\sum\limits_{k=1}^c\sum\limits_{x|a}\sum\limits_{y|b}\sum\limits_{z|c}[(x,y)=1][(y,z)=1][(x,z)=1]$

= \sum i = 1 a \sum j = 1 b \sum k = 1 c ⌊ a i ⌋ ⌊ b j ⌋ ⌊ c k ⌋ [(i, j) = 1] [(j, k) = 1] [(i, k) = 1]

$=\sum\limits_{i=1}^a\sum\limits_{j=1}^b\sum\limits_{k=1}^c{\lfloor{a\over i}\rfloor}{\lfloor {b\over j}\rfloor}{\lfloor {c\over k}\rfloor}[(i,j)=1][(j,k)=1][(i,k)=1]$
然后利用反演公式

[n=1]=∑d|nμ(d) $[n=1]=\sum\limits_{d|n}\mu(d)$ 将一个等式化开

= \sum i = 1 a \sum j = 1 b \sum k = 1 c ⌊ a i ⌋ ⌊ b j ⌋ ⌊ c k ⌋ \sum d | (i, j) μ (d) [(j, k) = 1] [(i, k) = 1]

$=\sum\limits_{i=1}^a\sum\limits_{j=1}^b\sum\limits_{k=1}^c{\lfloor{a\over i}\rfloor}{\lfloor {b\over j}\rfloor}{\lfloor {c\over k}\rfloor}\sum\limits_{d|(i,j)}\mu(d)[(j,k)=1][(i,k)=1]$

= \sum k = 1 c ⌊ c k ⌋ \sum d = 1 n μ (d) \sum i = 1 a [d | i] ⌊ a i ⌋ [(i, k) = 1] \sum j = 1 b [d | j] ⌊ b j ⌋ [(j, k) = 1]

$=\sum\limits_{k=1}^c{\lfloor {c\over k}\rfloor}\sum\limits_{d=1}^n\mu(d)\sum\limits_{i=1}^a[d|i]{\lfloor{a\over i}\rfloor}[(i,k)=1]\sum\limits_{j=1}^b[d|j]{\lfloor {b\over j}\rfloor}[(j,k)=1]$
令

i=di,j=dj $i=di,j=dj$

= \sum k = 1 c ⌊ c k ⌋ \sum d = 1 n μ (d) \sum i = 1 a d ⌊ a d i ⌋ [(i, k) = 1] \sum j = 1 b d ⌊ b d j ⌋ [(j, k) = 1]

$=\sum\limits_{k=1}^c{\lfloor {c\over k}\rfloor}\sum\limits_{d=1}^n\mu(d)\sum\limits_{i=1}^{a\over d}{\lfloor{a\over di}\rfloor}[(i,k)=1]\sum\limits_{j=1}^{b\over d}{\lfloor {b\over dj}\rfloor}[(j,k)=1]$
令

f(n,m)=∑i=1n⌊ni⌋[(i,m)=1] $f(n,m)=\sum\limits_{i=1}^n{\lfloor{n\over i}\rfloor}[(i,m)=1]$
那么原式

= \sum k = 1 c ⌊ c k ⌋ \sum d = 1 a μ (d) f (a d, k) f (b d, k)

$=\sum\limits_{k=1}^c{\lfloor {c\over k}\rfloor}\sum\limits_{d=1}^a\mu(d)f({a\over d},k) f({b\over d},k)$
求

f $f$ 是

O(n) $O(n)$ 的，那么总复杂度就是

O(n2logn) $O(n^2logn)$ 咯
优化常数的方法：gcd预处理/记忆化；因为abc的顺序是一定的无所谓的所以选两个小的枚举

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define N 2005
#define Mod 1073741824

int a,b,c,ans;
int p[N],prime[N],mu[N],G[N][N];

void get(int n)
{
    mu[1]=1;
    for (int i=2;i<=n;++i)
    {
        if (!p[i])
        {
            prime[++prime[0]]=i;
            mu[i]=-1;
        }
        for (int j=1;j<=prime[0]&&i*prime[j]<=n;++j)
        {
            p[i*prime[j]]=1;
            if (i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                break;
            }
            else mu[i*prime[j]]=-mu[i];
        }
    }
}
void Min(int &a,int &b,int &c)
{
    if (a>b) swap(a,b);
    if (b>c) swap(b,c);
    if (a>b) swap(a,b);
}
int gcd(int a,int b)
{
    if (!b) return a;
    else return gcd(b,a%b);
}
int F(int n,int m)
{
    int ans=0;
    for (int i=1;i<=n;++i)
        if (G[i][m]==1)
            ans+=n/i;
    return ans;
}
int main()
{
    get(2000);
    scanf("%d%d%d",&a,&b,&c);
    Min(a,b,c);
    for (int i=1;i<=c;++i)
        for (int j=i;j<=c;++j)
            G[i][j]=G[j][i]=gcd(i,j);
    for (int i=1;i<=a;++i)
        for (int j=1;j<=min(b,c);++j)
            if (G[i][j]==1)
                ans+=mu[j]*(a/i)*F(b/j,i)*F(c/j,i);
    ans=(ans%Mod+Mod)%Mod;
    printf("%d\n",ans);
}