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

题目描述

$T$组数据,给出$N$,$M$,求$sum_{x=1}^Nsum_{y=1}^M lim(x,y) ewline$
$N,M <= 10000000 ewline T<= 10000$

题目分析

直接开始变换,假设N<M
$Ans=sum_{x=1}^Nsum_{y=1}^M frac {xy}{(x,y)} ewline =sum_{T=1}^Nfrac 1Tsum_{x=1}^Nsum_{y=1}^Mxy[(x,y)==T] ewline =sum_{T=1}^Nfrac 1Tsum_{x=1}^{⌊frac NT⌋}sum_{y=1}^{⌊frac MT⌋}xyT^2[(x,y)==1] ewline =sum_{T=1}^NTsum_{x=1}^{⌊frac NT⌋}sum_{y=1}^{⌊frac MT⌋}xysum_{d|x,d|y}mu(d) ewline =sum_{d=1}^Nmu(d)sum_{T=1}^{N}Tsum_{d|x}^{⌊frac NT⌋}xsum_{d|y}^{⌊frac MT⌋}y ewline =sum_{d=1}^Nmu(d)sum_{T=1}^{N}Td^2sum_{x=1}^{⌊frac{⌊frac NT⌋}d⌋}xsum_{y=1}^{⌊frac{⌊frac MT⌋}d⌋}y ewline =sum_{d=1}^Nmu(d)sum_{T=1}^{N}Td^2sum_{x=1}^{⌊frac N{Td}⌋}xsum_{y=1}^{⌊frac M{Td}⌋}y ewline 此时令k=Td ewline Ans=sum_{k=1}^Nsum_{T|k}mu(⌊frac kT⌋)k⌊frac kT⌋sum_{x=1}^{⌊frac N{k}⌋}xsum_{y=1}^{⌊frac M{k}⌋}y ewline =sum_{k=1}^Nksum_{T|k}mu(T)Tsum_{x=1}^{⌊frac N{k}⌋}xsum_{y=1}^{⌊frac M{k}⌋}y ewline$
总算推完了…
此时只需要$Theta(N)$线性筛出$sum_{T|k}mu(T)T$,然后处理$ksum_{T|k}mu(T)T$的前缀和
而$sum_{x=1}^{⌊frac N{k}⌋}xsum_{y=1}^{⌊frac M{k}⌋}y$可以$Theta(1)$出
利用整除分块优化,时间复杂度为$Theta(N+Tsqrt N)$

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
", (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$Maxn$来预处理,否则会$TLE$,要读入$N$,$M$后再$O(N)$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
", (calc(N, M) + mod) % mod);
}