When Einstein stepped up the stairs, if every step was 2 steps, there were 1-order two ways.

Einstein came up with such a mathematical problem: there is a long ladder, if each step across 2, then the last one, if each step across 3, then the last 2 order, if each step across the 5-order, then the last 4 order, if each step across the 6-order, the last remaining 5 order. Only 7 steps at a time, the end is just the first order of no left. How many steps are there in this ladder?

Two ways:

Method 1: Experiment from 1 onwards

unsigned int  getstairnum ()
{for
	(unsigned int n = 1;; ++n)
	{
		if (1 = n% 2) && (2 = = n% 3) && (4 = n% 5) && (5 = n% 6) && (0 = n% 7)) return
			n;
	}

Or:

unsigned int  getstairnum ()  
{  
 unsigned int stairnum = 0;  
 int flag=1;  
 while (flag)  
 {  
  if (stairnum%2 ==1 && stairnum%3 ==2 && stairnum%5 = 4 && stairnum%6 = 5 && stairnum%7 = = 0)  
  {  
   flag =0;  
   return stairnum;  
  }   
  stairnum++;  
 }  
 return 0;  
}  

Method Two: Stair order is an odd number, and is a multiple of 7, so you can cross 14 per step (7+14 is odd, 7+7 is even), a loop

unsigned int  getstairnum ()
{
	int n = 7;
	while (1)
	{
		if ((1 = n% 2) && (2 = n% 3) && (4 = n% 5) && (5 = n% 6)) return
			n;< C7/>else
			n + +
	}
}

Or:

unsigned int  getstairnum ()
{
	unsigned int n = 7;
	while ((1!= n% 2) | | (2!= n% 3) | | (4!= n% 5) | | (5!= n% 6))
			n + +;
	return n;
}