Description
求 \(\sum_{i=1}^n\sum_{j=1}^m(n\%i)*(m\%j)\),\(i!=j\)
Solution
写成这样的形式：
\(\sum_{i=1}^{n}\sum_{j=1}^{m}(n-\lfloor\frac{n}{i}\rfloor*i)*(m-\lfloor\frac{m}{j}\rfloor*j)\)
暴力拆即可，注意 \(i=j\) 的情况，减去下式即可
\(\sum_{i=1}^{min(n,m)}(n\%i)*(m\%i)\)
模数不是质数，手算一个6的逆元即可
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=19940417;
int inv=3323403;
inline int S(int n){return 1ll*n*(n+1)/2%mod;}
inline int G(int n){return 1ll*n*(n+1)%mod*(2*n+1)%mod*inv%mod;}
void work()
{
int n,m;
cin>>n>>m;
if(n>m)swap(n,m);
ll ni=0,mj=0,rc=0;
for(int i=1,r;i<=n;i=r+1){
r=n/(n/i);
ni=(ni+1ll*(n/i)*(1ll*S(r)-S(i-1)+mod)%mod)%mod;
}
for(int j=1,r;j<=m;j=r+1){
r=m/(m/j);
mj=(mj+1ll*(m/j)*(1ll*S(r)-S(j-1)+mod)%mod)%mod;
}
for(int i=1,r;i<=n;i=r+1){
r=min(n/(n/i),m/(m/i));
rc=(rc+1ll*(n/i)*(m/i)%mod*(1ll*G(r)-G(i-1)+mod)%mod)%mod;
}
rc=(rc+1ll*n*n%mod*m%mod-ni*m%mod)%mod;
for(int i=1,r;i<=n;i=r+1){
r=min(n,m/(m/i));
rc=(rc-1ll*(m/i)*(1ll*S(r)-S(i-1)+mod)%mod*n%mod)%mod;
}
ll ans=1ll*n*n%mod*m%mod*m%mod;
ans=ans-mj*n%mod*n%mod-ni*m%mod*m%mod+ni*mj%mod-rc;
ans%=mod;if(ans<0)ans+=mod;
printf("%lld\n",ans);
}
int main()
{
freopen("pp.in","r",stdin);
freopen("pp.out","w",stdout);
work();
return 0;
}