[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$\sum_{x=1}^N\sum_{y=1}^M lim(x,y)\newline$

题目分析

直接开始变换,假设N<M
$Ans=\sum_{x=1}^N\sum_{y=1}^M \frac {xy}{(x,y)}\newline =\sum_{T=1}^N\frac 1T\sum_{x=1}^N\sum_{y=1}^Mxy[(x,y)==T]\newline =\sum_{T=1}^N\frac 1T\sum_{x=1}^{⌊\frac NT⌋}\sum_{y=1}^{⌊\frac MT⌋}xyT^2[(x,y)==1]\newline =\sum_{T=1}^NT\sum_{x=1}^{⌊\frac NT⌋}\sum_{y=1}^{⌊\frac MT⌋}xy\sum_{d|x,d|y}\mu(d)\newline =\sum_{d=1}^N\mu(d)\sum_{T=1}^{N}T\sum_{d|x}^{⌊\frac NT⌋}x\sum_{d|y}^{⌊\frac MT⌋}y\newline =\sum_{d=1}^N\mu(d)\sum_{T=1}^{N}Td^2\sum_{x=1}^{⌊\frac{⌊\frac NT⌋}d⌋}x\sum_{y=1}^{⌊\frac{⌊\frac MT⌋}d⌋}y\newline =\sum_{d=1}^N\mu(d)\sum_{T=1}^{N}Td^2\sum_{x=1}^{⌊\frac N{Td}⌋}x\sum_{y=1}^{⌊\frac M{Td}⌋}y\newline 此时令k=Td\newline Ans=\sum_{k=1}^N\sum_{T|k}\mu(⌊\frac kT⌋)k⌊\frac kT⌋\sum_{x=1}^{⌊\frac N{k}⌋}x\sum_{y=1}^{⌊\frac M{k}⌋}y\newline =\sum_{k=1}^Nk\sum_{T|k}\mu(T)T\sum_{x=1}^{⌊\frac N{k}⌋}x\sum_{y=1}^{⌊\frac M{k}⌋}y\newline$
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$k\sum_{T|k}\mu(T)T$的前缀和
而$\sum_{x=1}^{⌊\frac N{k}⌋}x\sum_{y=1}^{⌊\frac M{k}⌋}y$可以$\Theta(1)$出
利用整除分块优化,时间复杂度为$\Theta(N+T\sqrt N)$

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\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$Maxn$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\n", (calc(N, M) + mod) % mod);
}