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.
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.
Print the number of prime-factor prime numbers in [l,r].
#include<bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
typedef long long ll;
const int maxn=1e6+5;
int prime[maxn],v[40005],cnt,num[maxn],str[maxn];//num是保存因子个数，str是lr区间里面的数
int vis[38]= {0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1};//判断因子个数是否素数
void getp(int n)//得到sqrt（r）内的素数，线性筛
{
for(register int i=2; i<=n; ++i)
{
if(v[i]==0)
v[i]=i,prime[++cnt]=i;
for(register int j=1; j<=cnt; ++j)
{
if(prime[j]>v[i]||prime[j]>n/i)
break;
v[i*prime[j]]=prime[j];
}
}
}
int main()
{
int l,r,templ,tempr,poi;
scanf("%d %d",&l,&r);
int len=r-l+1;
getp(sqrt(r));
for(register int i=l; i<=r; ++i)//赋值
str[i-l+1]=i;
for(register int j=1; j<=cnt; ++j)//对sqrt(r)内的素数进行枚举
{
templ=ceil(l/(prime[j]*1.0));//i的左区间边际
tempr=floor(r/(1.0*prime[j]));//i的右区间边际
for(register int i=templ; i<=tempr; ++i)//枚举i
{
poi=i*prime[j]-l+1;
if(poi>len)continue;
while(str[poi]%prime[j]==0)//除去因子
str[poi]=str[poi]/prime[j],++num[poi];
}
}
int ans=0;
for(register int i=1; i<=len; ++i)//当str[i]还不为1时说明质因子个数还得加1，因为不为1的数肯定剩下的是素数
if((str[i]==1&&vis[num[i]])||(str[i]!=1&&vis[num[i]+1]))++ans;
printf("%d\n",ans);
return 0;
}
