P3768 【简单的数学题】
\(Ans=\sum ^{n}_{i=1}\sum ^{n}_{j=1}ijgcd(i,j)\)###
\(=\sum ^{n}_{i=1}\sum ^{n}_{j=1}ij\sum _{k|i,k|j} φ(k)\)###
\(=\sum ^{n}_{k=1} φ(k) \sum _{k|i}\sum _{k|j}ij\)###
\(=\sum ^{n}_{k=1}\varphi (k) k^{2} (\sum ^{n/k}_{i=1}i)^{2}\)###
\(=\sum ^{n}_{k=1}\varphi (k) k^{2} \sum ^{n/k}_{i=1}i^{3}\)###
\(n<=1e10\)杜教筛筛 \(\varphi (k) k^{2}\)
#include<bits/stdc++.h>
#include<tr1/unordered_map>
using namespace std;
typedef long long LL;
const int maxn=5000000+9;
inline LL Read(){
LL x=0,f=1; char c=getchar();
while(c<'0'||c>'9'){
if(c=='-') f=-1; c=getchar();
}
while(c>='0'&&c<='9')
x=(x<<3)+(x<<1)+c-'0',c=getchar();
return x*f;
}
LL n,p,inv;
int phi[maxn],prime[maxn];
LL sum[maxn];
bool visit[maxn];
inline LL Pow(LL base,LL b){
LL a=1;
while(b){
if(b&1)
a=(a*base)%p;
base=(base*base)%p;
b>>=1;
}
return a;
}
inline void F_phi(int N){
inv=Pow(6,p-2);
phi[1]=1; int tot(0);
for(int i=2;i<=N;++i){
if(!visit[i]){
phi[i]=i-1;
prime[++tot]=i;
}
for(int j=1;j<=tot&&i*prime[j]<=N;++j){
visit[i*prime[j]]=true;
if(i%prime[j]==0){
phi[i*prime[j]]=phi[i]*prime[j];
break;
}
else
phi[i*prime[j]]=phi[i]*phi[prime[j]];
}
}
for(int i=1;i<=N;++i)
sum[i]=(sum[i-1]+1ll*i*i%p*phi[i]%p)%p;
//for(int i=1;i<=2000;++i)
// printf("%lld ",sum[i]);printf("\n");
}
tr1::unordered_map<LL,LL> w;
inline LL S2(LL x){
x%=p;
return x*(x+1)%p*(2*x+1)%p*inv%p;
}
LL S3(LL x){
x%=p;
return (x*(x+1)/2)%p*((x*(x+1)/2)%p)%p;
}
LL Calc(LL now){
if(now<=5000000)
return sum[now];
if(w[now])
return w[now];
LL num=S3(now);
for(LL l=2,r;l<=now;l=r+1){
r=now/(now/l);
num=(num-Calc(now/l)*(S2(r)-S2(l-1)+p)%p+p)%p;
}
return w[now]=num;
}
int main(){
p=Read(),n=Read();
F_phi(5000000);
LL ans(0);
for(LL l=1,r;l<=n;l=r+1){
r=n/(n/l);
ans=(ans+S3(n/l)*((Calc(r)-Calc(l-1)+p)%p))%p;
}
printf("%lld",ans);
return 0;
}/*
998244353 2000
883968974
*/
