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.
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.
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\n",ans); 72 return 0; 73 } 74 /* 75 1 9 76 77 78 100000000 101000000 79 */
