[BZOJ3561]DZY Loves Math VI
Description
给定正整数n,m。求
\[\sum\limits_{i=1}^n\sum\limits_{j=1}^mlcm(i,j)^{\gcd(i,j)}
\]
Input
一行两个整数n,m。
Output
一个整数，为答案模1000000007后的值。
Sample Input
5 4
Sample Output
424
HINT
数据规模：
1<=n,m<=500000，共有3组数据。
首先推柿子
\[\sum\limits_{i=1}^n\sum\limits_{j=1}^mlcm(i,j)^{\gcd(i,j)}
\]
\[\sum\limits_{i=1}^n\sum\limits_{j=1}^m(\dfrac{i\times j}{\gcd(i,j)})^{\gcd(i,j)}
\]
\[\sum\limits_{d=1}^n\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}(\dfrac{d^2\times i\times j}{d})^d[\gcd(i,j)=1]
\]
\[\sum\limits_{d=1}^nd^d\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}i^d\times j^d\sum\limits_{x|i,x|j}\mu(x)
\]
\[\sum\limits_{d=1}^nd^d\sum\limits_{x=1}^{\lfloor\frac{n}{d}\rfloor}\mu(x)\times x^{2d}\sum\limits_{i=1}^{\lfloor\frac{n}{dx}\rfloor}i^d\sum\limits_{j=1}^{\lfloor\frac{m}{dx}\rfloor}j^d
\]
然后设\(T=dx\)，然后就。。。布星，推不下去了
发现\(m\leqslant 5×10^5\)，那不就得了，直接这样求就好，枚举d的时候，维护一下\(f(n)=\sum\limits_{i=1}^n i^d\)即可，复杂度约为\(O(m\log m)\)
/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline int read(){
int x=0,f=1;char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x>=10) print(x/10);
putchar(x%10+'0');
}
const int N=5e5,p=1e9+7;
int prime[N+10],miu[N+10];
bool inprime[N+10];
void prepare(){
int tot=0; miu[1]=1;
for (int i=2;i<=N;i++){
if (!inprime[i]) prime[++tot]=i,miu[i]=-1;
for (int j=1;j<=tot&&prime[j]*i<=N;j++){
inprime[i*prime[j]]=1;
if (i%prime[j]==0) break;
miu[i*prime[j]]=-miu[i];
}
}
}
int mlt(int a,int b){
int res=1;
for (;b;b>>=1,a=1ll*a*a%p) if (b&1) res=1ll*res*a%p;
return res;
}
int val[N+10],sum[N+10];
int main(){
prepare();
int n=read(),m=read(),Ans=0;
if (n>m) swap(n,m);
for (int i=1;i<=m;i++) val[i]=1;
for (int d=1;d<=n;d++){
int x=mlt(d,d),res=0,limn=n/d,limm=m/d;
for (int i=1;i<=limm;i++) val[i]=1ll*val[i]*i%p,sum[i]=(sum[i-1]+val[i])%p;
for (int i=1;i<=limn;i++){
if (!miu[i]) continue;
res=(res+1ll*val[i]*val[i]%p*sum[limn/i]%p*sum[limm/i]%p*miu[i]+p)%p;
}
Ans=(Ans+1ll*x*res%p)%p;
}
printf("%d\n",Ans);
return 0;
}
