除数函数求和 2 [LOJ125]-CFANZ编程社区

传送门

$Ans=\sum _{i=1}^n\sum _{d|i}(2*d^2+3*d+5)=\sum_{d=1}^n(2*d^2+3*d+5)[n/d]$

$\sum _{i=1}^ni^2=n*(n+1)*(2n+1)/6$

然后整除分块

#include<bits/stdc++.h>
#define Mod 998244353
#define LL long long
using namespace std;
int n;
LL Sum(LL x){return x * (x+1) % Mod * (x*2+1) % Mod * 166374059 % Mod;}
LL calc(LL r,LL l){
  LL val1 = Sum(r) - Sum(l-1);
  LL val2 = (l+r) * (r-l+1) % Mod * 499122177 % Mod;
  LL val3 = r-l+1;
  return (val1 * 2 + val2 * 3 + val3 * 5) % Mod;
}
int main(){
  scanf("%d",&n); LL ans = 0;
  for(int l=1,r;l<=n;l=r+1){
    int val = n/l; r = n/val;
    ans += calc((LL)r,(LL)l) * (LL)val % Mod;
    ans = (ans % Mod + Mod) % Mod;
  } printf("%lld",ans); return 0;
}