题解 | #KiKi求质数个数#
KiKi求质数个数
https://www.nowcoder.com/practice/d3a404ee0f8d41f98bf4707035d91086
#include <math.h>
#include <stdio.h>
int main()
{
int count = 0;
for (int i = 101; i <= 997; i += 2)
{
int j = 2;
for (j = 3; j <= sqrt(i); j++)
{
if (i % j == 0)
{
break;
}
}
if ((j * j) > i)
{
count++;
}
}
printf("%d", count);
return 0;
}
