【BZOJ】2956: 模积和

题意

求\(\sum_{i=1}^{n} \sum_{j=1}^{m} (n \ mod \ i)(m \ mod \ j)[i \neq j] \ mod \ 19940417\), \((n, m \le 10^9)\)

分析

以下均设\(n \le m\)

$$
\begin{align}
&
\sum_{i=1}^{n} \sum_{j=1}^{m} (n \ mod \ i)(m \ mod \ j)[i \neq j] \ mod \ 19940417
\\

\equiv &

\left(

\sum_{i=1}^{n}

\sum_{j=1}^{m}

(n \ mod \ i)(m \ mod \ j)

\sum_{i=1}^{n}

(n \ mod \ i \cdot m \ mod \ i)

\right)

\ mod \ 19940417

\

\equiv &

\left(

\left(

\sum_{i=1}^{n}

(n \ mod \ i)

\right)

\left(

\sum_{j=1}^{m}

(m \ mod \ i)

\right)

\sum_{i=1}^{n}

(n \ mod \ i \cdot m \ mod \ i)

\right)

\ mod \ 19940417

\

\end{align}

\[</p>

于是我们只需要快速求出$\sum_{i=1}^{n} ( n \ mod \ i)$和$\sum_{i=1}^{n} ( n \ mod \ i \cdot m \ mod \ i )$就能解决问题了。

<p>
\]

\begin{align}

& \sum_{i=1}^{n} ( n \ mod \ i)

\

= &

\sum_{i=1}^{n}

\left( n - i \left \lfloor \frac{n}{i} \right \rfloor \right)

\

= &

n^2

\sum_{i=1}^{n}

i \left \lfloor \frac{n}{i} \right \rfloor

\

& \sum_{i=1}^{n} ( n \ mod \ i \cdot \ m \ mod \ i)

\

= &

\sum_{i=1}^{n}

\left( n - i \left \lfloor \frac{n}{i} \right \rfloor \right) \left( m - i \left \lfloor \frac{m}{i} \right \rfloor \right)

\

= &

n^2m

+

\sum_{i=1}^{n}

i^2 \left \lfloor \frac{n}{i} \right \rfloor \left \lfloor \frac{m}{i} \right \rfloor

n\sum_{i=1}^{n}

i \left \lfloor \frac{m}{i} \right \rfloor

m\sum_{i=1}^{n}

i \left \lfloor \frac{n}{i} \right \rfloor

\

\end{align}

\[</p>

## 题解
于是分块大法好...

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mo=19940417;
ll cal(int n, ll a) {
ll ret=a%mo*n%mo, tp=0;
for(int i=1, pos=0; i<=n; i=pos+1) {
pos=n/(n/i);
tp+=(a/i)%mo*(((ll)(pos+1)*pos/2-(ll)(i-1)*i/2)%mo)%mo;
if(tp>=mo) {
tp-=mo;
}
}
return (ret-tp+mo)%mo;
}
int main() {
int n, m;
scanf("%d%d", &n, &m);
if(n>m) {
swap(n, m);
}
printf("%lld\n", (cal(n, n)*cal(m, m)%mo-cal(n, (ll)n*m)+mo)%mo);
return 0;
}\]