POJ 3978(求素数)

2014-11-24 13:06:02 · 作者: · 浏览: 0
知识点: 1.求素数的test,从2~sqrt(n); 2.如果数据很多,可以用素数表记录,然后sum=prime[m]-prime[n]求得!!!!!!!!!!!!

Primes
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 3204 Accepted: 1245


Description

A pretty straight forward task, calculate the number of primes between 2 integers.

Given 2 integers A ≤ B < 10 5 what’s the number of primes in range from A to B inclusive.

Note: A prime number is a positive integer greater than 1 and is divisible by 1 and itself only. For N to be prime it is enough to test the divisibility of numbers less than or equal to square root of N.

Input

As many as 1000 lines, each line contains 2 integers A and B separated by a space. Input is terminated when A = B = -1 (Do not process this line).

Output

For every line in input except for the last line where A = B = -1 - print the number of prime numbers between A and B inclusive.

Sample Input

0 9999
1 5
-1 -1

Sample Output

1229
3


#include
  
   
#include
   
     #include
    
      using namespace std; bool is_prime(int n) { if(n<=0||n==1) //贡献1 WA,a,b可能小于0;so n<=0 return false; return false; for(int i=2;i<=sqrt((double)n);i++) { if(n%i==0) return false; } return true; } int main() { int a,b; int sum; while(1) { sum=0; scanf("%d%d",&a,&b); if(a==-1&&b==-1) break; for(int i=a;i<=b;i++) if(is_prime(i)) ++sum; printf("%d\n",sum); } return 0; }