【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 "
"
#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+7;
ll qpow(ll x, ll y){ll s=1;while(y){if(y&1)s=s*x%mod;x=x*x%mod;y>>=1;}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=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();return x*f;}
  
const int maxn = 1e5+5;
ll vis[maxn], prime[maxn], cnt;
void isprime()
{
    memset(vis,0,sizeof(vis));
    for(ll i=2;i<maxn;i++)
    {
        if(!vis[i])prime[cnt++]=i;
        for(ll j=0;j<cnt && i*prime[j]<maxn;j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)break;
        }
    }
    vis[1] = 1;
    vis[0] = 1;
}
  
const int N = 1e6+5;
ll bas[N];
int amo[N];
  
int main()
{
    int l, r;
    isprime() ;
  
    scanf("%d%d",&l,&r);
    for(int i=0;i<=r-l+5;i++) bas[i] = 1;
    for(int i=0;i<cnt;i++)
    {
        int u = (l-1)/prime[i] + 1;
        int v = r/prime[i];
        for(int j=u;j<=v;j++)
        {
            int _ = j*prime[i], __=_;
            while(_%prime[i]==0){
                amo[__-l+1]++;
                _ /= prime[i];
                bas[__-l+1] *= prime[i];
            }
        }
    }
    //cout<<1<<nl;
    int ans = 0;
    for(int i=l;i<=r;i++)
    {
        if(i<=3) continue;
        //cout<<i<<" : "<<amo[i-l+1]<<nl;
        if(bas[i-l+1] != i)amo[i-l+1] ++;
        if(!vis[amo[i-l+1]]) ans ++;
    }
    cout<<ans<<nl;
  
    return 0;
}