整除分块
一般形式：\(\sum_{i = 1}^n \lfloor \frac{n}{i} \rfloor * f(i)\)。
需要一种高效求得函数 \(f(i)\) 的前缀和的方法，比如等差等比数列求和或对于积性函数的筛法等，然后就可以用整除分块的思想做。
题目解法
化公式变成比较方便的形式：
\(\ \sum_{i = 1}^n \sum_{j = 1}^m (n \mod i)(m \mod j), i \ne j\)
\(= \sum_{i = 1}^n \sum_{j = 1}^m (n - i \lfloor \frac{n}{i} \rfloor)(m - j \lfloor \frac{m}{j} \rfloor) - \sum_{i = 1}^{min(n, m)} (n - i \lfloor \frac{n}{i} \rfloor)(m - i \lfloor \frac{m}{i} \rfloor)\)
乘法分配律展开，化简，令 \(t = min(n, m)\) 得：
\(\ \sum_{i = 1}^n \sum_{j = 1}^m (n - i \lfloor \frac{n}{i} \rfloor)(m - j \lfloor \frac{m}{j} \rfloor) - \sum_{i = 1}^t (n - i \lfloor \frac{n}{i} \rfloor)(m - i \lfloor \frac{m}{i} \rfloor)\)
\(= \sum_{i = 1}^n \sum_{i = 1}^m nm + m\sum_{i = 1}^n \sum_{i = 1}^m j \lfloor \frac{n}{i} \rfloor + n\sum_{i = 1}^n \sum_{i = 1}^m \lfloor \frac{m}{j} \rfloor +\)
\(\sum_{i = 1}^n \sum_{i = 1}^m ij \lfloor \frac{n}{i} \rfloor \lfloor \frac{m}{j} \rfloor - \sum_{i = 1}^t nm + m \sum_{i = 1}^t i \lfloor \frac{n}{i} \rfloor - n \sum_{i = 1}^t i \lfloor \frac{m}{i} \rfloor - \sum_{i = 1}^t i^2 \lfloor \frac{n}{i} \rfloor\)
\(\sum_{i = 1}^n \sum_{i = 1}^m ij \lfloor \frac{n}{i} \rfloor \lfloor \frac{m}{j} \rfloor\) 等于 \(\sum_{i = 1}^n i \lfloor \frac{n}{i} \rfloor * \sum_{i = 1}^m j \lfloor \frac{m}{j} \rfloor\)，一个一个算即可。
代码写的比较长……因为用 \(unsigned ll\) 为了避免出负数也多了很多取模……
#include <queue>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef unsigned long long u64;
const u64 mod = 19940417;
const u64 inv_6 = 3323403;
inline u64 Calc_1(u64 l, u64 r) { return (l + r) * (r - l + 1) / 2 % mod; }
inline u64 Calc_2(u64 x) { return ((x + 1) * (2 * x + 1) % mod) * (x * inv_6 % mod) % mod; }
int main(int argc, const char *argv[])
{
u64 n = 0, m = 0, t = 0, ans = 0, sum_1 = 0, sum_2 = 0;
scanf("%llu%llu", &n, &m);
ans = ((n * m % mod) * (n * m % mod) % mod);
t = min(n, m), ans = (ans + mod - (n * m % mod) * t % mod) % mod;
for(u64 tmp, l = 1, r = 1; l <= n; l = r + 1) {
tmp = n / l, r = n / tmp;
ans = (ans + mod - (Calc_1(l, r) * tmp % mod) * (m * m % mod) % mod) % mod;
sum_1 = (sum_1 + Calc_1(l, r) * tmp % mod) % mod;
}
for(u64 tmp, l = 1, r = 1; l <= m; l = r + 1) {
tmp = m / l, r = m / tmp;
ans = (ans + mod - (Calc_1(l, r) * tmp % mod) * (n * n % mod) % mod) % mod;
sum_2 = (sum_2 + Calc_1(l, r) * tmp % mod) % mod;
}
for(u64 tmp, l = 1, r = 1; l <= t; l = r + 1) {
tmp = n / l, r = min(t, n / tmp);
ans = (ans + Calc_1(l, r) * (tmp * m % mod)) % mod;
}
for(u64 tmp, l = 1, r = 1; l <= t; l = r + 1) {
tmp = m / l, r = min(t, m / tmp);
ans = (ans + Calc_1(l, r) * (tmp * n % mod)) % mod;
}
for(u64 l = 1, r = 1; l <= t; l = r + 1) {
r = min(t, min(n / (n / l), m / (m / l)));
ans = (ans + mod - (mod + Calc_2(r) - Calc_2(l - 1)) * ((n / l) * (m / l) % mod) % mod) % mod;
}
printf("%llu\n", (ans + sum_1 * sum_2) % mod);
return 0;
}原文地址：https://www.cnblogs.com/nanjoqin/p/10261968.html