CF#235E. Number Challenge

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可以理解为上一道题的扩展板..

然后我们就可以YY出这样一个式子

${\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^cd(ijk)=\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^c[gcd(i,j)=gcd(i,k)=gcd(j,k)=1]\lfloor\frac{a}{i}\rfloor\lfloor\frac{b}{j}\rfloor\lfloor\frac{c}{k}\rfloor}$

然后我们枚举第一维，排除掉不和第一维互质的数大力反演就OK啦。

这里介绍另一种很神奇（麻烦）方法，当然这个和反演的关系就没那么大了：

首先设$f(k)=\sum\limits_{i=1}^{a}\sum\limits_{j=1}^{b}[k=i \times j]$

然后得到$ans=\sum\limits_{i=1}^{a\times b} f(i)\sum\limits_{j=1}{d(i \times j)}$

$ans=\sum\limits_{i=1}^{a\times b} f(i)\sum\limits_{j=1}^{c}{d(i \times j)}$
$ans=\sum_{i=1}^{ab}f(i)\sum_{j=1}^c\sum_{u|i}\sum_{v|j}[gcd(u,v)=1]$
$ans=\sum\limits_{i=1}^{a\times b}\sum\limits_{j=1}^{c}[gcd(i,j)=1]\sum\limits_{u=1}^{\frac{a\times b}{i}}f(u\times i)\lfloor \frac{c}{j}\rfloor$

不妨设$S(i)=\sum\limits_{u=1}^{\frac{a\times b}{i}}f(u\times i)$

得到：

$ans=\sum\limits_{i=1}^{a\times b}\sum\limits_{j=1}^{c}[gcd(i,j)=1]S(i)\lfloor \frac{c}{j}\rfloor$

$ans=\sum\limits_{i=1}^{a\times b}\sum\limits_{j=1}^{c}\sum\limits_{d|gcd(i,j)}\mu(d)S(i)\lfloor \frac{c}{j}\rfloor$

不妨设$Q(k)=\sum\limits_{i=1}^{k}\lfloor \frac{k}{i} \rfloor$

得到：

$ans=\sum\limits_{i=1}^{c}\mu(i)\sum\limits_{j=1}^{\frac{a\times b}{i}}S(i\times j)\sum\limits_{k=1}^{\frac{c}{i}}Q(k)$

不妨设$P(k)=\sum\limits_{i=1}^{\frac{a\times b}{k}}S(i\times k)$

最后得到$ans=\sum\limits_{i=1}^{c}\mu(i)P(i)Q(i)$

大力预处理即可。

//CF235E
//by Cydiater
//2017.2.22
#include <iostream>
#include <queue>
#include <map>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <iomanip>
#include <algorithm>
#include <bitset>
#include <set>
#include <vector>
#include <complex>
using namespace std;
#define ll int
#define up(i,j,n)	for(ll i=j;i<=n;i++)
#define down(i,j,n)	for(ll i=j;i>=n;i--)
#define cmax(a,b)	a=max(a,b)
#define cmin(a,b)	a=min(a,b)
const ll MAXN=4e6+5;
const ll mod=1073741824;
inline ll read(){
	char ch=getchar();ll x=0,f=1;
	while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
ll A,B,C,AB,prime[MAXN],P[MAXN],S[MAXN],f[MAXN],Q[MAXN],miu[MAXN],cnt;
bool vis[MAXN];
namespace solution{
	void Prepare(){
		A=read();B=read();C=read();AB=A*B;
		miu[1]=1;
		up(i,2,C){
			if(!vis[i]){prime[++cnt]=i;miu[i]=-1;}
			up(j,1,cnt){
				if(i*prime[j]>C)break;
				vis[i*prime[j]]=1;
				if(i%prime[j])miu[i*prime[j]]=-miu[i];
				else break;
			}
		}
		up(i,1,A)up(j,1,B)f[i*j]++;
		up(i,1,AB)for(ll j=i;j<=AB;j+=i)
			(S[i]+=f[j])%=mod;
		up(i,1,AB)for(ll j=i;j<=AB;j+=i)
			(P[i]+=S[j])%=mod;
		ll pos;
		up(i,1,C){
			for(ll j=1;j<=i;j=pos+1){
				pos=i/(i/j);
				Q[i]+=(pos-j+1)*(i/j);
				(Q[i]+=mod)%=mod;
			}
		}
	}
	void Solve(){
		ll ans=0;
		up(i,1,C){
			(ans+=miu[i]*P[i]*Q[C/i])%=mod;
			(ans+=mod)%=mod;
		}
		cout<<ans<<endl;
	}
}
int main(){
	//freopen("input.in","r",stdin);
	using namespace solution;
	Prepare();
	Solve();
	//cout<<"Time has passed:"<<1.0*clock()/1000<<"s!"<<endl;
	return 0;
}