int SumOfPrime(int n)
{
int i, k, sum = 0;
for (i = 2; i < n; i++)
{
bool prime = true;
for (k = 2; k <= i / 2; k++)
{
if (i % k == 0)
{
prime = false;
break;
}
}
if (prime)
sum += i;
}
return sum;
}
计算n以内(不包含n)的所有质数之和
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转载自blog.csdn.net/huangziguang/article/details/104435803
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