【数学】Prime-Factor Prime

Prime-Factor Prime

题目描述

A positive integer is called a "prime-factor prime" when the number of its prime factors is prime. For example, 12 is a prime-factor prime because the number of prime factors of 12=2×2×3 is 3, which is prime. On the other hand, 210 is not a prime-factor prime because the number of prime factors of 210=2×3×5×7 is 4, which is a composite number.

In this problem, you are given an integer interval [l,r]. Your task is to write a program which counts the number of prime-factor prime numbers in the interval, i.e. the number of prime-factor prime numbers between l and r, inclusive.

输入

The input consists of a single test case formatted as follows.

l r
A line contains two integers l and r (1≤l≤r≤109), which presents an integer interval [l,r]. You can assume that 0≤r−l<1,000,000.

输出

Print the number of prime-factor prime numbers in [l,r].

样例输入

1 9

样例输出

4

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【题解】

区间素数筛即可解决。

【队友代码】

 //
 //  IniAully
 //
 //#pragma GCC optimize("Ofast,no-stack-protector")
 //#pragma GCC optimize("O3")
 #pragma GCC optimize(2)
 #include <bits/stdc++.h>
 #define inf 0x3f3f3f3f
 #define linf 0x3f3f3f3f3f3f3f3fll
 #define pi acos(-1.0)
 #define nl "\n"
 #define pii pair<ll,ll>
 #define ms(a,b) memset(a,b,sizeof(a))
 #define FAST_IO ios::sync_with_stdio(NULL);cin.tie(NULL);cout.tie(NULL)
 using namespace std;
 typedef long long ll;
 const ll mod = 1e9+;
 ll qpow(ll x, ll y){ll s=;while(y){if(y&)s=s*x%mod;x=x*x%mod;y>>=;}return s;}
 //ll qpow(ll a, ll b){ll s=1;while(b>0){if(b%2==1)s=s*a;a=a*a;b=b>>1;}return s;}
 inline int read(){int x=,f=;char ch=getchar();while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}while(ch>=''&&ch<='') x=x*+ch-'',ch=getchar();return x*f;}

 const int maxn = 1e5+;
 ll vis[maxn], prime[maxn], cnt;
 void isprime()
 {
     memset(vis,,sizeof(vis));
     for(ll i=;i<maxn;i++)
     {
         if(!vis[i])prime[cnt++]=i;
         for(ll j=;j<cnt && i*prime[j]<maxn;j++)
         {
             vis[i*prime[j]]=;
             if(i%prime[j]==)break;
         }
     }
     vis[] = ;
     vis[] = ;
 }

 const int N = 1e6+;
 ll bas[N];
 int amo[N];

 int main()
 {
     int l, r;
     isprime() ;

     scanf("%d%d",&l,&r);
     for(int i=;i<=r-l+;i++) bas[i] = ;
     for(int i=;i<cnt;i++)
     {
         int u = (l-)/prime[i] + ;
         int v = r/prime[i];
         for(int j=u;j<=v;j++)
         {
             int _ = j*prime[i], __=_;
             while(_%prime[i]==){
                 amo[__-l+]++;
                 _ /= prime[i];
                 bas[__-l+] *= prime[i];
             }
         }
     }
     //cout<<1<<nl;
     int ans = ;
     for(int i=l;i<=r;i++)
     {
         if(i<=) continue;
         //cout<<i<<" : "<<amo[i-l+1]<<nl;
         if(bas[i-l+] != i)amo[i-l+] ++;
         if(!vis[amo[i-l+]]) ans ++;
     }
     cout<<ans<<nl;

     return ;
 }