POJ 3978（求素数）

知识点：

     1.求素数的test，从2~sqrt（n）。

    

     2.假设数据非常多。能够用素数表记录。然后sum=prime[m]-prime[n]求得！！！。！

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                                                                                             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 < 10⁵ 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<iostream>
#include<cstdio>
#include<cmath>

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
",sum);

    }
    return 0;
}