[bzoj 2693] jzptab & [bzoj 2154] Crash的数字表格 (莫比乌斯反演)

题目描述

TTT组数据,给出NNN,MMM,求∑x=1N∑y=1Mlim(x,y)\sum_{x=1}^N\sum_{y=1}^M lim(x,y)\newlinex=1∑N​y=1∑M​lim(x,y)

N,M<=10000000T<=10000N,M <= 10000000\newline
T&lt;= 10000N,M<=10000000T<=10000

题目分析

直接开始变换,假设N<M

Ans=∑x=1N∑y=1Mxy(x,y)=∑T=1N1T∑x=1N∑y=1Mxy[(x,y)==T]=∑T=1N1T∑x=1⌊NT⌋∑y=1⌊MT⌋xyT2[(x,y)==1]=∑T=1NT∑x=1⌊NT⌋∑y=1⌊MT⌋xy∑d∣x,d∣yμ(d)=∑d=1Nμ(d)∑T=1NT∑d∣x⌊NT⌋x∑d∣y⌊MT⌋y=∑d=1Nμ(d)∑T=1NTd2∑x=1⌊⌊NT⌋d⌋x∑y=1⌊⌊MT⌋d⌋y=∑d=1Nμ(d)∑T=1NTd2∑x=1⌊NTd⌋x∑y=1⌊MTd⌋y此时令k=TdAns=∑k=1N∑T∣kμ(⌊kT⌋)k⌊kT⌋∑x=1⌊Nk⌋x∑y=1⌊Mk⌋y=∑k=1Nk∑T∣kμ(T)T∑x=1⌊Nk⌋x∑y=1⌊Mk⌋y
Ans=\sum_{x=1}^N\sum_{y=1}^M \frac {xy}{(x,y)}\newline
=\sum_{T=1}^N\frac 1T\sum_{x=1}^N\sum_{y=1}^Mxy[(x,y)==T]\newline
=\sum_{T=1}^N\frac 1T\sum_{x=1}^{⌊\frac NT⌋}\sum_{y=1}^{⌊\frac MT⌋}xyT^2[(x,y)==1]\newline
=\sum_{T=1}^NT\sum_{x=1}^{⌊\frac NT⌋}\sum_{y=1}^{⌊\frac MT⌋}xy\sum_{d|x,d|y}\mu(d)\newline
=\sum_{d=1}^N\mu(d)\sum_{T=1}^{N}T\sum_{d|x}^{⌊\frac NT⌋}x\sum_{d|y}^{⌊\frac MT⌋}y\newline
=\sum_{d=1}^N\mu(d)\sum_{T=1}^{N}Td^2\sum_{x=1}^{⌊\frac{⌊\frac NT⌋}d⌋}x\sum_{y=1}^{⌊\frac{⌊\frac MT⌋}d⌋}y\newline
=\sum_{d=1}^N\mu(d)\sum_{T=1}^{N}Td^2\sum_{x=1}^{⌊\frac N{Td}⌋}x\sum_{y=1}^{⌊\frac M{Td}⌋}y\newline
此时令k=Td\newline
Ans=\sum_{k=1}^N\sum_{T|k}\mu(⌊\frac kT⌋)k⌊\frac kT⌋\sum_{x=1}^{⌊\frac N{k}⌋}x\sum_{y=1}^{⌊\frac M{k}⌋}y\newline
=\sum_{k=1}^Nk\sum_{T|k}\mu(T)T\sum_{x=1}^{⌊\frac N{k}⌋}x\sum_{y=1}^{⌊\frac M{k}⌋}y\newline
Ans=x=1∑N​y=1∑M​(x,y)xy​=T=1∑N​T1​x=1∑N​y=1∑M​xy[(x,y)==T]=T=1∑N​T1​x=1∑⌊TN​⌋​y=1∑⌊TM​⌋​xyT2[(x,y)==1]=T=1∑N​Tx=1∑⌊TN​⌋​y=1∑⌊TM​⌋​xyd∣x,d∣y∑​μ(d)=d=1∑N​μ(d)T=1∑N​Td∣x∑⌊TN​⌋​xd∣y∑⌊TM​⌋​y=d=1∑N​μ(d)T=1∑N​Td2x=1∑⌊d⌊TN​⌋​⌋​xy=1∑⌊d⌊TM​⌋​⌋​y=d=1∑N​μ(d)T=1∑N​Td2x=1∑⌊TdN​⌋​xy=1∑⌊TdM​⌋​y此时令k=TdAns=k=1∑N​T∣k∑​μ(⌊Tk​⌋)k⌊Tk​⌋x=1∑⌊kN​⌋​xy=1∑⌊kM​⌋​y=k=1∑N​kT∣k∑​μ(T)Tx=1∑⌊kN​⌋​xy=1∑⌊kM​⌋​y

总算推完了…

此时只需要Θ(N)\Theta(N)Θ(N)线性筛出∑T∣kμ(T)T\sum_{T|k}\mu(T)T∑T∣k​μ(T)T,然后处理k∑T∣kμ(T)Tk\sum_{T|k}\mu(T)Tk∑T∣k​μ(T)T的前缀和

而∑x=1⌊Nk⌋x∑y=1⌊Mk⌋y\sum_{x=1}^{⌊\frac N{k}⌋}x\sum_{y=1}^{⌊\frac M{k}⌋}y∑x=1⌊kN​⌋​x∑y=1⌊kM​⌋​y可以Θ(1)\Theta(1)Θ(1)出

利用整除分块优化,时间复杂度为Θ(N+TN)\Theta(N+T\sqrt N)Θ(N+TN​)

AC code([bzoj 2693] jzptab)

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 10000005, mod = 1e8+9;
int N, M;
namespace Mobius
{
	int mu[MAXN], Prime[MAXN], cnt;
	bool IsnotPrime[MAXN];
	int sum[MAXN];
	void init()
	{
		sum[1] = 1;
		for(int i = 2; i <= MAXN-5; i++)
		{
			if(!IsnotPrime[i]) Prime[++cnt] = i, sum[i] = 1-i;
			for(int j = 1; j <= cnt && i * Prime[j] <= MAXN-5; j++)
			{
				IsnotPrime[i * Prime[j]] = 1;
				if(i % Prime[j] == 0) { sum[i * Prime[j]] = sum[i]; break; }
				sum[i * Prime[j]] = 1ll * sum[i] * (1 - Prime[j]) % mod;
			}
		}
		for(int i = 1; i <= MAXN-5; i++)//前缀和
			sum[i] = (sum[i-1] + 1ll*sum[i]*i%mod) % mod;
	}
	int Sum(int N, int M)
	{
		return ((1ll*N*(N+1)/2) % mod) * ((1ll*M*(M+1)/2) % mod) % mod;
	}
	int calc(int N, int M)
	{
		int ret = 0;
		for(int i = 1, j; i <= N; i=j+1)//整除分块
		{
			j = min(N/(N/i), M/(M/i));
			ret = (ret + 1ll * (sum[j] - sum[i-1]) % mod * Sum(N/i, M/i) % mod) % mod;
		}
		return ret;
	}
}
using namespace Mobius;
int main ()
{
	int T; init();
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d%d", &N, &M); if(N > M) swap(N, M);
		printf("%d\n", (calc(N, M) + mod) % mod);
	}
}

AC code([bzoj 2154] Crash的数字表格)

这道题有个恶心的地方,不能用MaxnMaxnMaxn来预处理,否则会TLETLETLE,要读入NNN,MMM后再O(N)O(N)O(N)处理

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 10000005, mod = 20101009;
int N, M;
namespace Mobius
{
	int mu[MAXN], Prime[MAXN], cnt;
	bool IsnotPrime[MAXN];
	int sum[MAXN];
	void init()
	{
		sum[1] = 1;
		for(int i = 2; i <= N; i++)
		{
			if(!IsnotPrime[i]) Prime[++cnt] = i, sum[i] = 1-i;
			for(int j = 1; j <= cnt && i * Prime[j] <= N; j++)
			{
				IsnotPrime[i * Prime[j]] = 1;
				if(i % Prime[j] == 0) { sum[i * Prime[j]] = sum[i]; break; }
				sum[i * Prime[j]] = 1ll * sum[i] * (1 - Prime[j]) % mod;
			}
		}
		for(int i = 1; i <= N; i++)
			sum[i] = (sum[i-1] + 1ll*sum[i]*i%mod) % mod;
	}
	int Sum(int N, int M)
	{
		return ((1ll*N*(N+1)/2) % mod) * ((1ll*M*(M+1)/2) % mod) % mod;
	}
	int calc(int N, int M)
	{
		int ret = 0;
		for(int i = 1, j; i <= N; i=j+1)
		{
			j = min(N/(N/i), M/(M/i));
			ret = (ret + 1ll * (sum[j] - sum[i-1]) % mod * Sum(N/i, M/i) % mod) % mod;
		}
		return ret;
	}
}
using namespace Mobius;
int main ()
{
	scanf("%d%d", &N, &M); if(N > M) swap(N, M); init();
	printf("%d\n", (calc(N, M) + mod) % mod);
}