【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}

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