素数个数
求a,b之间的素数个数
每行输入a和b（0<=a,b<=500000）
每行输出包括a和b在内的素数个数
Sample Input
0 3
5 296
9 9
Sample Output
2
60
0
[解决办法]
#include <iostream>
#include <algorithm>
using namespace std;
bool prime(int Number);
int main( )
{
while(true)
{
int a,b,sum=0;
cout<<"please input two number:";
cin >> a >> b;
if(max(a,b)<=50000&&0<=min(a,b)){
int i;
if(min(a,b)<2)
{
i=2;
}
else
{
i=min(a,b);
}
for(i; i <= max(a,b); i++)
{
if (prime(i))
{
sum++;
}
}
cout<<sum<<endl;
}
}
return 0;
}
bool prime(int Number)
{
//（在此补充）
int i;
for(i=2;i*i<=Number;++i)
if(Number%i==0)
return false;
return true;
}
[解决办法]
#include <stdio.h>int isprime(unsigned long n){ unsigned long i; if(n == 1 || n == 2 || n == 3 || n == 5) { return 0; } else if(n % 2) { for(i = 3; i <= n / 2 + 1; i += 2) { if(n % i == 0) { return -1; } } return 0; } else { return -1; }}unsigned long count_prime(unsigned long m, unsigned long n){ unsigned count = 0; unsigned long i; for(i = m; i <= n; i++) { if(isprime(i) == 0) { count++; } } return count;}int main(int argc, char* argv[]){ unsigned long m; unsigned long n; do{ scanf("%lu %lu", &m, &n); printf("%lu\n", count_prime(m, n)); }while(m != 0 || n != 0); return 0;}