Question

求大神们用C#帮我解一下这道题

Answer 1

//判断数字中是否存在7
private static bool isSeven(int num)
{
int count = 0,temp;
bool bFlage = false;
while (num!=0)
{
temp = num % 10;
if (temp == 7)
{
count++;
}
num /= 10;
}
return bFlage = count > 0 ? true : false;
}
static void Main(string[] args)
{
int count = 0,countA = 0;
int[] num = new int[1000];
Console.WriteLine("1-1000的恶魔数字有：");
for (int i = 1; i <= 1000; i++)
{
if (isSeven(i) == true)
{
num[count] = i;
count++;
}
else if (isSeven(i) == false && i % 7 == 0)
{
num[count] = i;
count++;
}
}
for (int i = 0; i < count; i++)
{
countA++;
if (countA % 17 == 0 || countA == count)
{
Console.WriteLine(num[i].ToString());
}
else
{
Console.Write(string.Format("{0},",num[i]));
}
}
Console.ReadLine();
}

Answer 2

程序参考：手机打的没法去测试了，目测差不多

检查一个数各位是否有7
bool CheckSeven(int num)
{
while(num!=0)
{
if(i%10==7) return true;
num=num/10;
}
return false;
}

判断是否7的倍数
bool CheckTimes(int num)
{
return num%7==0;
}

主函数
int main()
{
for(int i=0;i<=1000;i++)
{
if(CheckSeven(i)||CheckTimes(i)) Console.WriteLine(i);
}
}