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Description:
Count the number of prime numbers less than a non-negative number, n
Hint: The number n could be in the order of 100,000 to 5,000,000.
#define NO 0
#define YES 1
class Solution {
public:
int countPrimes(int n) {
if(n == 1)
return 0;
int num;
int count = 0;
for(num = 2;num <= n;num++){
if(isPrime(num))
count++;
}
return count;
}
int isPrime(int n){
int i = 2;
for(;i<=sqrt(n);i++)
{
if( (n%i) == 0)
return NO;
}
return YES;
}
};
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原文地址：http://www.cnblogs.com/xuanyuanchen/p/4460952.html
