BZOJ3994: [SDOI2015]约数个数和

Description

设d(x)为x的约数个数，给定N、M，求

[sum_{i=1}^{n}sum_{j=1}^md(i*j) ]

Input

输入文件包含多组测试数据。

第一行，一个整数T，表示测试数据的组数。

接下来的T行，每行两个整数N、M。

Output

T行，每行一个整数，表示你所求的答案。

Sample Input

2
7 4
5 6

Sample Output

110
121

HINT

1<=N, M<=50000

1<=T<=50000

Solution

做这题首先要知道约数函数(d(x))的一个性质，不然完全做不下去
注：((x,y))表示(gcd(x,y))

[large d(i*j)=sum_{x|i}sum_{y|j}[(x,y)=1] ]

证明的话..不难理解，但是写一遍很麻烦，所以直接看这篇文章吧...
https://blog.csdn.net/ab_ever/article/details/76737617

所以我们就可以开始推式子了

[large{ egin{aligned} &sum_{i=1}^{n}sum_{j=1}^md(i*j)\ &=sum_{i=1}^{n}sum_{j=1}^msum_{x|i}sum_{y|j}[(x,y)=1]\ &=sum_{x=1}^{n}sum_{y=1}^{m}sum_{i=1}^{lfloorfrac{n}{x} floor}sum_{j=1}^{lfloorfrac{m}{y} floor}[(x,y)=1]\ &=sum_{x=1}^{n}sum_{y=1}^{m}lfloorfrac{n}{x} floorlfloorfrac{m}{y} floor[(x,y)=1]\ &=sum_{x=1}^{n}sum_{y=1}^{m}lfloorfrac{n}{x} floorlfloorfrac{m}{y} floorsum_{d|(x,y)}mu(d)\ &=sum_{d=1}^{n}sum_{x=1}^{lfloorfrac{n}{d} floor}sum_{y=1}^{lfloorfrac{m}{d} floor}lfloorfrac{n}{dx} floorlfloorfrac{m}{dy} floor*mu(d)\ &=sum_{d=1}^{n}mu(d)sum_{x=1}^{lfloorfrac{n}{d} floor}sum_{y=1}^{lfloorfrac{m}{d} floor}lfloorfrac{n}{dx} floorlfloorfrac{m}{dy} floor\ &=sum_{d=1}^{n}mu(d)sum_{x=1}^{lfloorfrac{n}{d} floor}lfloorfrac{n}{dx} floorsum_{y=1}^{lfloorfrac{m}{d} floor}lfloorfrac{m}{dy} floor\ end{aligned}\ 设g(x)=sum_{i=1}^{n}{lfloorfrac{n}{i} floor}\ 则代入原式可得\ egin{aligned} &sum_{d=1}^{n}mu(d)sum_{x=1}^{lfloorfrac{n}{d} floor}lfloorfrac{n}{dx} floorsum_{y=1}^{lfloorfrac{m}{d} floor}lfloorfrac{m}{dy} floor\ &=sum_{d=1}^{n}mu(d)*g(lfloorfrac{n}{d} floor)*g(lfloorfrac{m}{d} floor) end{aligned} } ]

那么处理g的函数值和莫比乌斯函数的函数值，就可以(O(Tsqrt{n}))处理了

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 50010
int p[N], vis[N], mu[N], sum[N];
int n, m, cnt;
ll ans = 0, g[N];

void init() {
	mu[1] = 1;
	for(int i = 2; i < N; ++i) {
		if(!vis[i]) p[++cnt] = i, mu[i] = -1;
		for(int j = 1; j <= cnt && i * p[j] < N; ++j) {
			vis[i * p[j]] = 1;
			if(i % p[j] == 0) break;
			mu[i * p[j]] -= mu[i];
		}
	}
	for(int i = 1; i < N; ++i) {
		sum[i] = sum[i - 1] + mu[i];
		for(int l = 1, r; l <= i; l = r + 1) {
			r = i / (i / l);
			g[i] += 1ll * (r - l + 1) * (i / l);
		}
	}
}

int main() {
	init();
	int T;
	scanf("%d", &T);
	while(T--) {
		scanf("%d%d", &n, &m);
		if(n > m) swap(n, m);
		ans = 0;
		for(int l = 1, r; l <= n; l = r + 1) {
			r = min(n/(n/l), m/(m/l));
			ans += 1ll * (sum[r] - sum[l - 1]) * g[n / l] * g[m / l];
//			printf("%d %d %d
", (sum[r] - sum[l - 1]), g[n / l], g[m / l]);
		}
		printf("%lld
", ans);
	}
	return 0;
}