【清华集训2012】模积和

Description

给定n,m，求 $\sum\limits_{i=1}^{n} \! \sum\limits_{j=1}^m[i\neq j](n\  mod \  i)(m \  mod \  j)\mod 19940417$的值

Solution

我们先求出$\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}(n\ mod\ i)(m\ mod\ j)$的值，然后运用容斥原理排除i=j的情况即可。

$$ans=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}(n\ mod\ i)(m\ mod\ j)-\sum\limits_{i=1}^{min(n,m)}(n\ mod\ i)(m\ mod\ i)$$
$$=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}(n-\lfloor\frac{n}{i}\rfloor i)(m-\lfloor\frac{m}{j}\rfloor j)-\sum\limits_{i=1}^{min(n,m)}(n-\lfloor\frac{n}{i}\rfloor i)(m-\lfloor\frac{m}{i}\rfloor i)$$
$$=\sum\limits_{i=1}^{n}(n-\lfloor\frac{n}{i}\rfloor i)\sum\limits_{j=1}^{m}(m-\lfloor\frac{m}{j}\rfloor j)-\sum\limits_{i=1}^{min(n,m)}(nm+\lfloor\frac{n}{i}\rfloor\lfloor\frac{m}{i}\rfloor i^2-(m\lfloor\frac{n}{i}\rfloor+n\lfloor\frac{m}{i}\rfloor)i)$$

另外，本题中我们需要一个引理：$\sum\limits_{i=1}^{n}{i^2}=\frac{n(n+1)(2n+1)}{6}$

这样，我们预处理6的乘法逆元，然后根据上述推导用整除分块计算即可

Code

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ll;
ll n, m;
const ll mod = 19940417;
inline ll sum1(ll x, ll y) {
    return (x + y) * (y - x + 1) / 2 % mod;
}
inline ll sum2(ll x) {
    return (x * (x + 1)) % mod * (x + x + 1) % mod * 3323403 % mod;
}
ll calc(ll x) {
    ll ret = 0;
    for (register ll l = 1, r; l <= x; l = r + 1) {
        r = x / (x / l);
        ret = (ret + (r - l + 1) * x % mod - (x / l) * sum1(l, r) % mod + mod) % mod; 
    }
    return ret;
}
int main() {
    scanf("%lld%lld", &n, &m);
    ll ans = 0;
    ans = calc(n) * calc(m) % mod;
    for (register ll l = 1, r; l <= min(n, m); l = r + 1) {
        r = min(n / (n / l), m / (m / l));
        ll s1 = (n * m % mod * (r - l + 1)) % mod;
        ll s2 = (n / l) * (m / l) % mod * (sum2(r) - sum2(l - 1) + mod) % mod;
        ll s3 = (n / l * m % mod + m / l * n % mod) * sum1(l, r) % mod;
        ans = (ans - (s1 + s2 - s3 + mod) % mod + mod) % mod; 
    }
    printf("%lld\n", ans % mod);
    return 0;
}