#include<bits/stdc++.h>
using namespace std;
const int N=1e7+5,mod=20101009;
bool NotPrime[N];
int tot;
int Prime[N],f[N];
long long pre[N];
void sieve(int n){
NotPrime[1]=1;
f[1]=1;
for(int i=2;i<=n;i++){
if(!NotPrime[i]){
Prime[tot++]=i;
f[i]=1-i;
}
for(int j=0,k;(k=i*Prime[j])<=n;j++){
NotPrime[k]=1;
if(i%Prime[j]==0){
f[k]=f[i];
break;
}
f[k]=f[i]*f[Prime[j]];
}
}
for(int i=1;i<=n;i++){
pre[i]=(pre[i-1]+(long long)f[i]*i%mod+mod)%mod;
}
}
int main(){
int n,m;
scanf("%d%d",&n,&m);
if(n>m)swap(n,m);
sieve(n);
long long ans=0;
for(int i=1,j;i<=n;i=j+1){
j=min(n/(n/i),m/(m/i));
long long s1=(long long)(1+n/i)*(n/i)/2%mod,s2=(long long)(1+m/i)*(m/i)/2%mod;
ans+=s1%mod*s2%mod*(pre[j]-pre[i-1]+mod)%mod;
ans%=mod;
}
printf("%lld\n",ans);
return 0;
}
线性筛原理解释:
题目原始式子是\(\sum_{i=1}^n\)
P1829
原文:https://www.cnblogs.com/-9-QAQ-6-/p/13045559.html