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 < 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.
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\n",sum);
}
return 0;
}