BZOJ 3930 [CQOI2015]选数

题目链接

https://lydsy.com/JudgeOnline/problem.php?id=3930

题解

反演得到
$\sum_{i=1}^{\lfloor h/k\rfloor} \mu(i) (\lfloor \frac{h}{ki}\rfloor-\lfloor\frac{l-1}{ki}\rfloor)^n$
整除分块直接算即可。

代码

#include <map>
#include <cstdio>
#include <algorithm>

int read()
{
  int x=0,f=1;
  char ch=getchar();
  while((ch<'0')||(ch>'9'))
    {
      if(ch=='-')
        {
          f=-f;
        }
      ch=getchar();
    }
  while((ch>='0')&&(ch<='9'))
    {
      x=x*10+ch-'0';
      ch=getchar();
    }
  return x*f;
}

const int maxn=1000000;
const int mod=1000000007;
const int inf=0x3f3f3f3f;

int p[maxn+10],prime[maxn+10],cnt,mu[maxn+10];

int getprime()
{
  p[1]=mu[1]=1;
  for(int i=2; i<=maxn; ++i)
    {
      if(!p[i])
        {
          prime[++cnt]=i;
          mu[i]=mod-1;
        }
      for(int j=1; (j<=cnt)&&(i*prime[j]<=maxn); ++j)
        {
          int x=i*prime[j];
          p[x]=1;
          if(i%prime[j]==0)
            {
              mu[x]=0;
              break;
            }
          mu[x]=mod-mu[i];
        }
    }
  for(int i=1; i<=maxn; ++i)
    {
      mu[i]+=mu[i-1];
      if(mu[i]>=mod)
        {
          mu[i]-=mod;
        }
    }
  return 0;
}

std::map<int,int> mp;

int getsum(int n)
{
  if(n<=maxn)
    {
      return mu[n];
    }
  if(mp.count(n))
    {
      return mp[n];
    }
  int ans=1;
  for(int l=2,r; l<=n; l=r+1)
    {
      r=n/(n/l);
      ans-=1ll*(r-l+1)*getsum(n/l)%mod;
      if(ans<0)
        {
          ans+=mod;
        }
    }
  return mp[n]=ans;
}

int n,k,s,t;

int quickpow(int a,int b)
{
  int res=1;
  while(b)
    {
      if(b&1)
        {
          res=1ll*res*a%mod;
        }
      a=1ll*a*a%mod;
      b>>=1;
    }
  return res;
}

int main()
{
  getprime();
  n=read();
  k=read();
  s=read();
  t=read();
  int ans=0;
  for(int l=1,r; l<=t/k; l=r+1)
    {
      r=inf;
      if(t/(l*k)!=0)
        {
          r=std::min(r,t/(t/(l*k)));
        }
      if((s-1)/(l*k)!=0)
        {
          r=std::min(r,(s-1)/((s-1)/(l*k)));
        }
      r/=k;
      ans=(ans+1ll*(getsum(r)-getsum(l-1)+mod)*quickpow((t/(k*l))-((s-1)/(k*l)),n))%mod;
    }
  printf("%d\n",ans);
  return 0;
}