【BZOJ2693】jzptab & 【BZOJ2154】Crash的数字表格

题目

弱化版题目的传送门（【BZOJ2154】Crash的数字表格）

加强版题目的传送门（【BZOJ2693】jzptab）

思路&解法

题目是要求： (sumlimits_{i = 1}^{n}sumlimits_{j = 1}^{m}lcm(i, j))

于是我们可以把式子化成这样：

[sum_{i = 1}^{n}sum_{j = 1}^{m}frac{ij}{gcd(i, j)} ]

然后我们枚举gcd

[sum_{i = 1}^{n}sum_{j = 1}^{m} sum_{k = 1}^{min(n, m)}frac{ij}{k}[gcd(i, j) == k] ]

我们再把式子换一下

[sumlimits_{k = 1}^{min(n, m)}{frac{1}{k}sumlimits_{i = 1}^{n}sumlimits_{j = 1}^{m}ij[gcd(i, j) == k]} ]

[sumlimits_{k = 1}^{min(n, m)}{frac{1}{k}sumlimits_{i = 1}^{lfloor{frac{n}{k}} floor}sumlimits_{j = 1}^{lfloorfrac{m}{k} floor}ijk^2[gcd(i, j) == 1]} ]

[sum^{min(n, m)}_{k = 1}ksum_{i= 1}^{lfloor{frac{n}{k}} floor}sum_{j = 1}^{lfloor{frac{m}{k}} floor}ij[gcd(i, j) ==1] ]

反演一下

[sum_{k}^{min(n, m)} k sum_{d = 1}^{min(lfloor {frac{n}{k}} floor, lfloor {frac{m}{k}} floor)}mu(d) imes sum_{i|d}sum_{j|d} ij ]

[sum_{k}^{min(n, m)} ksum_{d = 1}^{min(lfloor {frac{n}{k}} floor, lfloor {frac{m}{k}} floor)} mu(d) imes d^2sum_{i=1}^{lfloor{frac{n}{kd}} floor}sum_{j=1}^{lfloorfrac{m}{kd} floor} ij ]

[sum_{k}^{min(n, m)} ksum_{d = 1}^{min(lfloor {frac{n}{k}} floor, lfloor {frac{m}{k}} floor)} mu(d)d^2 F(lfloor{frac{n}{kd}} floor, lfloor{frac{m}{kd}} floor) ]

其中$$F(n, m) = {nm(n+1)(m+1)over 4}$$

继续优化

[sum_{T=1}^{min(n, m)}{F(lfloor{frac{n}{T}} floor, lfloor{frac{m}{T}} floor)}sum_{d|T}mu(d)d^2{frac{T}{d}} ]

后面的(sumlimits_{d|T}mu(d)d^2{frac{T}{d}})的前缀和很容易求

代码

【BZOJ2693】jzptab

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const LL mod = 100000009LL;

const int N = 10000010;

int p[N], total;
bool vis[N];

LL g[N];

void init()
{	g[1] = 1;
	for (int i = 2; i <= 10000000; i++)
	{	if (!vis[i]) p[++total] = i, g[i] = (LL) (1 - i + mod) % mod;
		for (int j = 1; j <= total && i * (LL) p[j] <= 10000000; j++)
		{	vis[i * p[j]] = 1;
			if (i % p[j] == 0) { g[i * p[j]] = g[i]; break; }
			else g[i * p[j]] = (g[i] * g[p[j]]) % mod;
		}
	}
	for (int i = 2; i <= 10000000; i++) g[i] = (g[i] * i + g[i-1]) % mod;
}

LL Get(int n) { return ((LL) n * (LL) (n+1) / 2LL) % mod; }

LL Sum(int n, int m) { return (Get(n) * Get(m)) % mod; }

LL Calc(int n, int m)
{	int last = 0;
	LL Ans = 0;
	for (int i = 1; i <= min(n, m); i = last+1)
	{	last = min(n / (n/i), m / (m/i));
		Ans = (Ans + (Sum(n/i, m/i) * (g[last] - g[i-1])) % mod) % mod;
	}
	return (Ans + mod) % mod;
}

int main()
{	init();
	int T;
	scanf("%d", &T);
	while (T--)
	{	int n, m;
		scanf("%d %d", &n, &m);
		printf("%lld
", Calc(n, m));
	}
	return 0;
}

【BZOJ2154】Crash的数字表格

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const LL mod = 20101009LL;

const int N = 10000010;

int p[N], total;
bool vis[N];

LL g[N];

void init()
{	g[1] = 1;
	for (int i = 2; i <= 10000000; i++)
	{	if (!vis[i]) p[++total] = i, g[i] = (LL) (1 - i + mod) % mod;
		for (int j = 1; j <= total && i * (LL) p[j] <= 10000000; j++)
		{	vis[i * p[j]] = 1;
			if (i % p[j] == 0) { g[i * p[j]] = g[i]; break; }
			else g[i * p[j]] = (g[i] * g[p[j]]) % mod;
		}
	}
	for (int i = 2; i <= 10000000; i++) g[i] = (g[i] * i + g[i-1]) % mod;
}

LL Get(int n) { return ((LL) n * (LL) (n+1) / 2LL) % mod; }

LL Sum(int n, int m) { return (Get(n) * Get(m)) % mod; }

LL Calc(int n, int m)
{	int last = 0;
	LL Ans = 0;
	for (int i = 1; i <= min(n, m); i = last+1)
	{	last = min(n / (n/i), m / (m/i));
		Ans = (Ans + (Sum(n/i, m/i) * (g[last] - g[i-1])) % mod) % mod;
	}
	return (Ans + mod) % mod;
}

int main()
{	init();
	int n, m;
	scanf("%d %d", &n, &m);
	printf("%lld
", Calc(n, m));
	return 0;
}

一些其他的东西

弱化版题目可以(O(n))过， 然而我是用(O(sqrt{n}))的算法做的， 而且达到了惊人的18s， 比(O(n))的解法慢多了。我哪里写锉了。。。。。。