习题2.6 1000以内自然数中3的倍数之和为

int main() {
    int x = 3;
    int sum = 0;
    while (x<=1000) {
        sum+= x;
        x+= 3;
    }
    printf("1000以内自然数中3的倍数之和为：%d",sum);
    return 0;

}

//for循环
int main()
{
	int sum = 0, i;
	for (i = 0; i < 1000; i++)
		if (i % 3 == 0)
			sum += i;
	printf("和为：%dn", sum);
	return 0;
}