洛谷 P2260 [清华集训2012]模积和

恶心至极！！！！！！！！

思路

求 (sumlimits_{i = 1}^{n} (n mod i)sumlimits_{j=1}^{m}(mmod j)[i eq j])

假设没有限制情况(i eq j)

(sumlimits_{i = 1}^{n} (n mod i)sumlimits_{j=1}^{m}(mmod j))

只看左半部分：

( sumlimits_{i=1}^{n}(n\% i))

(= sumlimits_{i=1}^{n}(n - lfloor frac{n}{i} floor * i))

(= sumlimits_{i=1}^{n}n - sumlimits_{i = 1}^{n}lfloorfrac{n}{i} floor*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

$ sumlimits_{i=1}^{k=min(n,m)}(nmod i)(mmod i)$

(=sumlimits_{i=1}^{k}(n-lfloorfrac{n}{i} floor*i)(m-lfloorfrac{m}{i} floor*i))

(=sumlimits_{i=1}^{k}(nm-(lfloorfrac{n}{i} floor*m+lfloorfrac{m}{i} floor*n)*i+lfloorfrac{n}{i} floor*lfloorfrac{m}{i} floor*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

   [sumlimits_{i = l}^{r}i=(r -l + 1)*(l + r) / 2 ]
   [sumlimits_{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;
}