bzoj 2956: 模积和

Description

求 (sum_{i=1}^nsum_{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-lfloorfrac{n}{i} floor*i)*(m-lfloorfrac{m}{j} floor*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
",ans);
}
int main()
{
	freopen("pp.in","r",stdin);
	freopen("pp.out","w",stdout);
	work();
	return 0;
}