JAG Asia 2017 C-----Prime-Factor Prime(素数区间筛)

C: Prime-Factor Prime

时间限制: 1 Sec 内存限制: 128 MB

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
题目大意：
给一个区间[l,r],求区间中满足素因子个数也是素数的数的个数。
12=2×2×3 ------ 3个素因子，3是素数，所以12满足条件
210=2×3×5×7 --- 4个素因子，4不是素数，所以210不满足条件

AC代码：

 1 /*
 2  *只有合数才可能满足题目要求
 3  *任何一个合数n必定包含一个不超过sqrt（n）的素因子
 4  */
 5 
 6 
 7 #include <bits/stdc++.h>
 8 using namespace std;
 9 typedef long long ll;
10 const int maxn=33000;
11 int prime[maxn+1];
12 bool isPrime[maxn+1];
13 void get_prime(int num)
14 {
15     memset(isPrime,false,sizeof(isPrime));
16     memset(prime,0,sizeof(prime));
17     for(int i=2; i<=num; ++i)
18     {
19         if(!prime[i])prime[++prime[0]]=i,isPrime[i]=true;
20         for(int j=1; j<=prime[0]&&prime[j]<=num/i; ++j)
21         {
22             prime[prime[j]*i]=1;
23             if(i%prime[j]==0)break;
24         }
25     }
26 }
27 int arr[1000007];//存区间数
28 int num[1000007];//存区间数的素因子个数
29 int main()
30 {
31     int a,b;
32     scanf("%d%d",&a,&b);
33     int len=b-a+1;
34     for(int i=1; i<=len; ++i)
35     {
36         arr[i]=a+i-1;
37     }
38     int m=ceil(sqrt(b));//必须向上取整，否则会出错39     get_prime(m);
40     int ans=0;
41     for(int i=1; i<=prime[0]; ++i)
42     {
43         int lef=ceil(a*1.0/prime[i]);
44         int rig=b*1.0/prime[i];
45         for(int j=lef; j<=rig; ++j)
46         {
47             /*
48              *只有合数才可能满足题目要求
49              *j*prime[i]必为合数
50              */
51             int pos=j*prime[i]-a+1;
52             while(arr[pos]%prime[i]==0)
53             {
54                 arr[pos]/=prime[i];
55                 ++num[pos];
56             }
57         }
58     }
59 //    for(int i=1;i<=len;++i)
60 //    {
61 //        cout<<arr[i]<<endl;
62 //        //if((arr[i]==1&&isPrime[num[i]])||(arr[i]>1&&isPrime[num[i]+1]))++ans;
63 //    }
64 //    system("pause");
65     for(int i=1;i<=len;++i)
66     {
67         if((arr[i]==1&&isPrime[num[i]])||(arr[i]>1&&isPrime[num[i]+1]))++ans;
68         //arr[i]==1&&isPrime[num[i]]  的情况是素因子都小于sqrt（b）
69         //arr[i]>1&&isPrime[num[i]+1] 的情况就是有一个素因子是大于srtq（b）的，素因子个数直接加一就好了
70     }
71     printf("%d
",ans);
72     return 0;
73 }
74 /*
75 1 9
76 
77 
78 100000000 101000000
79 */