恶心至极！！！！！！！！

题目链接

思路

求 \(\sum\limits_{i = 1}^{n} (n \mod i)\sum\limits_{j=1}^{m}(m\mod j)[i\neq j]\)

假设没有限制情况\(i\neq j\)

\(\sum\limits_{i = 1}^{n} (n \mod i)\sum\limits_{j=1}^{m}(m\mod j)\)

只看左半部分：

\(\ \ \ \sum\limits_{i=1}^{n}(n\% i)\)

\(= \sum\limits_{i=1}^{n}(n - \lfloor \frac{n}{i}\rfloor * i)\)

\(= \sum\limits_{i=1}^{n}n - \sum\limits_{i = 1}^{n}\lfloor\frac{n}{i}\rfloor*i\)

显然数论分块，右半部分同理，都可以数论分块做(余数求和那道题的完全一样的做法)

再看\(i=j\)的情况，即

$ \sum\limits_{i=1}^{k=\min(n,m)}(n\mod i)(m\mod i)$

\(=\sum\limits_{i=1}^{k}(n-\lfloor\frac{n}{i}\rfloor*i)(m-\lfloor\frac{m}{i}\rfloor*i)\)

\(=\sum\limits_{i=1}^{k}(nm-(\lfloor\frac{n}{i}\rfloor*m+\lfloor\frac{m}{i}\rfloor*n)*i+\lfloor\frac{n}{i}\rfloor*\lfloor\frac{m}{i}\rfloor*i^2)\)

用数论分块求出上面两个式子，用总的减去下面这个式子，注意除法要用逆元

\(ps\)：

1. 能模就模，好事多模

2. 小知识点

   \[\sum\limits_{i = l}^{r}i=(r -l + 1)*(l + r) / 2 \]

   \[\sum\limits_{i = 1} ^{n}i^2= n * (n + 1) * (2* n + 1)/ 6 \]

时间复杂度\(O(\sqrt{n})\)

代码

/*
Author:loceaner
*/
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define int long long
using namespace std;

const int A = 5e5 + 11;
const int B = 1e6 + 11;
const int inv6 = 3323403;
const int inv2 = 9970209;
const int mod = 19940417;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

int n, m, k;

inline int sum(int l, int r) {
	return (r - l + 1) * (l + r) / 2 % mod;
}

inline int sum1(int r) {
	return r * (r + 1) % mod * (2 * r + 1) % mod * inv6 % mod;
}

inline int sum2(int l, int r) {
	return (sum1(r) - sum1(l - 1)) % mod;
}

inline int solve(int n) {
	int ans = n * n;
	for (int l = 1, r; l <= n; l = r + 1) {
		if (n / l == 0) break;
		r = min(n / (n / l), n);
		ans -= (n / l) % mod * sum(l, r) % mod;
		ans %= mod;
	}
	return (ans % mod + mod) % mod;
}

signed main() {
	n = read(), m = read();
	int ans1 = solve(n) * solve(m) % mod, ans2 = 0;
	for (int l = 1, r, now1, now2, now3, now4; l <= min(n, m); l = r + 1) {
		r = min(n / (n / l), m / (m / l));
		now1 = n * m % mod * (r - l + 1) % mod;
		now2 = ((n / l) * (m / l) % mod * sum2(l, r) % mod + mod) % mod;
		now3 = ((n / l) * m % mod + (m / l) * n % mod) * sum(l, r) % mod;
		now4 = (now1 + now2 - now3 + mod) % mod;
		ans2 = (ans2 + now4) % mod;
	}
	cout << ((ans1 - ans2) % mod + mod) % mod;
	return 0;
}

标签：lfloor,洛谷,limits,int,sum,P2260,模积,rfloor,mod
来源： https://www.cnblogs.com/loceaner/p/12731906.html