#include <stdio.h>
#include <math.h>
int main()
{
int i, j, k, line = 0, sum = 0;
for (i = 3; i < 100; i += 2)
{
k = sqrt(i * 1.0);
for (j = 2; j <= k; j++)
if (i % j == 0)
break;
if (j > k)
{
line++;
printf("%-2d ", i);
sum += i;
if (line % 5 == 0)
printf("\n");
}
}
printf("\n");
printf("它们的和为: %d\n", sum);
return 0;
}
运行效果:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/f31fbe096b63f624c4b899538444ebf81b4ca3fd?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)