There are n people in a circle, the order of automatic arranging, from the first person to start off (from the 1~3 count), where 3 of the people who reported out of the circle, asked the last person left in the first number.

         
* * Copyright (c) 2012, Computer College of Yantai University         
* All rights reserved.         
* Author:  Liu Tongbin       
* Date of Completion: December 05, 2012         
* Version number: v1.0  
* *               
Input Description:    
* Problem Description: There are n people in a circle, the order of automatic arranging, starting from the first person to count (from the 1~3 numbered), c9/>* who           report 3 exit Circle, ask the last person left in the first date.
* Program output:
* Problem Analysis: slightly        
* Algorithm design: Slightly * *

#include <iostream>

using namespace std;

int main ()
{
	int num[50];

	int i,j,k,m,n;

	int *p;

	cout<<endl<< "Please enter the total number:" <<endl;

	cin>>n;

	P=num;

	for (i=0;i<n;i++)
	{
		* (p+i) =i+1;   In order of 1 to n, each person is numbered
	}

	i=0;    I for each cycle of the count variable

	k=0;    K is the count variable m=0 according to 1 2 3 off

	;    M for the number of exits

	while (m<n-1)  //When the number of exits is less than n-1 (that is, the number of exits is greater than 1 o'clock) Execute loop body
	{
		if (* (p+i)!=0)
		{
			k++;
		}

		if (k==3)    //The exit number is set to 0
		{
			* (p+i) =0;

			k=0;

			m++;
		}

		i++;

		if (i==n)
		{
			i=0;//count to tail I revert to 0
		}
	} while

	(*p==0)
	{
		p++;
	}

	cout<< "The last one is the" <<*p<< "number. "<<endl;

	return 0;
}