洛谷P2260 [清华集训2012]模积和题解

题目链接：P2260 [清华集训2012]模积和

题意：

求
$\left(\sum_{i=1}^{n}\sum_{j=1}^{m}(n\bmod i)\times(m\bmod j),i\ne j \right) \mod 19940417$

根据容斥原理
$\sum_{i=1}^{n}\sum_{j=1}^{m}(n\bmod i)\times(m\bmod j),i\ne j \\= \sum_{i=1}^{n}\sum_{j=1}^{m}(n\bmod i)\times(m\bmod j) - \sum_{i=1}^{n}(n\bmod i)\times(m\bmod i)$
然后推推柿子
$\begin{aligned} &\sum_{i=1}^{n}\sum_{j=1}^{m}(n\bmod i)\times(m\bmod j) - \sum_{i=1}^{n}(n\bmod i)\times(m\bmod i) \\&=\sum_{i=1}^{n}\left(n-i\left\lfloor{\dfrac{n}{i}}\right\rfloor\right)\times\sum_{j=1}^{m}\left(m-j\left\lfloor{\dfrac{m}{j}}\right\rfloor\right) - \sum_{i=1}^{n}\left(n-i\left\lfloor{\dfrac{n}{i}}\right\rfloor\right)\times\left(m-i\left\lfloor{\dfrac{m}{i}}\right\rfloor\right) \\&=\left(n^2-\sum_{i=1}^{n}i\left\lfloor{\dfrac{n}{i}}\right\rfloor\right)\times\left(m^2-\sum_{i=1}^{m}i\left\lfloor{\dfrac{m}{i}}\right\rfloor\right)-\sum_{i=1}^{n}\left(nm-mi\left\lfloor{\dfrac{n}{i}}\right\rfloor-ni\left\lfloor{\dfrac{m}{i}}\right\rfloor+i^2\left\lfloor{\dfrac{n}{i}}\right\rfloor\left\lfloor{\dfrac{m}{i}}\right\rfloor\right) \\&=\left(n^2-\sum_{i=1}^{n}i\left\lfloor{\dfrac{n}{i}}\right\rfloor\right)\times\left(m^2-\sum_{i=1}^{m}i\left\lfloor{\dfrac{m}{i}}\right\rfloor\right)-n^2m+\sum_{i=1}^{n}\left(mi\left\lfloor{\dfrac{n}{i}}\right\rfloor+ni\left\lfloor{\dfrac{m}{i}}\right\rfloor-i^2\left\lfloor{\dfrac{n}{i}}\right\rfloor\left\lfloor{\dfrac{m}{i}}\right\rfloor\right) \end{aligned}$
有一个小的公式，这里就不证明了（逃
$\sum_{i=1}^{n}i^2 = \dfrac{n(n+1)(2n+1)}{6}$
然后到这里就没啥难度了

考虑数论分块

代码如下

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int p=19940417;
const int inv2=9970209;
const int inv6=3323403;
#define sum1(x) ((x)%p*(x+1)%p*inv2%p)
#define sum2(x) ((x)%p*(x+1)%p*(2*(x)%p+1)%p*inv6%p)
int solve(int x)
{
    int res=x%p*x%p;
    for(int l=1,r; l<=x; l=r+1)
    {
        r=x/(x/l);
        res=(res-(sum1(r)-sum1(l-1))%p*(x/l)%p)%p;
    }
    return res;
}
int n,m,a,b,c,ans;
signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    cin >> n >> m;
    if(n>m)swap(n,m);
    ans=(solve(n)*solve(m)%p-n%p*n%p*m%p)%p;
    for(int l=1,r; l<=n; l=r+1)
    {
        r=min(n/(n/l),m/(m/l));
        a=(a+(sum1(r)-sum1(l-1))%p*m%p*(n/l)%p)%p;
        b=(b+(sum1(r)-sum1(l-1))%p*n%p*(m/l)%p)%p;
        c=(c+(sum2(r)-sum2(l-1))%p*(n/l)%p*(m/l)%p)%p;
    }ans=((ans+a+b-c)%p+p)%p;
    cout << ans << endl;
    return 0;
}