Temper cattle ranking problem

Temper cattle sorting problem

Preface ：
**[ Low voice bb]：** Because it's a school exam ,, therefore ... I'm programming for test cases （ A little ~~） Because I'm not very good at , There is a problem with this code , It can only require the number of cattle to be less than 10, And the order of rotation cannot be greater than 2,,emm… And then there is DEV C++ May run incorrectly （ I don't know , Waiting for the boss' advice ...）, stay V.C 6.0 No problem .

but, I still smelled shamelessly , After all, I'm very happy .
so, be it so , I think my comments are quite clear .

/*  The procedure has great limitations , The number of cows must not exceed 10 only （ It can be modified manually ） , And the order of rotation can only be 1 Once or twice . */ 
#include<iostream>
using namespace std;
int main()
{
    
	// Enter the number of cattle and array  
	//i Is a variable that records the number of cycles  ,n Is array length  
	int a[10] = {
    0};
	int i,n;
	cout<<" Please enter the number of cows ：" ;
	scanf("%d",&n);
	cout<<" Please enter the temper value of cattle ："; 
	for (i=0;i<n;i++)
		scanf("%d",&a[i]);
		
	/*  Give rotation   use  count  Record the first value of each rotation  ,  Use a multiline array cun[jie][shu] To save the rotation value for the subsequent exchange process  ,  Use a single row array wei[weishu] To save the values that have appeared in the rotation   Used to judge whether the rotation is over  */
	int count = 1; 
	int lun[10];	
	int cun[10][10]={
    0};
	int cun1[10]={
    0};
	int wei[10];
	int weishu=0;
	int jie=0,shu =0;
    
	// Give rotation , Save the rotation results in  cun[jie][shu] And wei[weishu]  in . 
	while(1)
	{
    	 
		int i = 1;
		// Using the dead loop , Give the result of rotation in order . 
		while(1)
		{
    
			lun[0] = count ;
			lun[i] = a[lun[i-1]-1];	
			if(lun[1] == count)
				break;
			cun[jie][shu] = lun[i-1];
			wei[weishu] = lun[i-1];
			weishu++;shu++;
			
			if(lun[i] == count)
				break;
			i++;
		}
			
		// Yes wei[weishu] Search and check , Back to the  count  assignment 
		int s = 1;
		int jige = 0;
		for(s=1;s<=n;s++)
		{
    	
			int y=0;	
			int buxiangdeng=0;
			/* take  1  To  n  Each value of is related to  wei[weishu] Compare arrays  ,  If equal   Variable jige increase 1, Unequal variables buxiangdeng increase 1  When the cycle is complete  buxiangdeng = 10, Numbers s Not in  wei[weishu]  in   So the next one  count  Namely  s  if  s != 10 ,s It's just  wei[weishu]  There has been , It's not the next one  count  once  jige  Variables and  n  equal  , It means that all the figures are  wei[weishu]  There has been , End of rotation ！！*/ 
			for(y=0;y<10;y++)
			{
    
				if(s == wei[y])
				{
    
					jige++;	
				}	
				else
					buxiangdeng++;	
			}
			if(buxiangdeng ==10)
			{
    
				count =s;	break;		
			}
		}
		
		if(jige==n)
		{
    
			break;	
		}
		shu=0; jie++;
	}
	
	// Definition  n1  And  n2  Express   The number of numbers in each rotation  
	int n1=0,n2=0;
	for(i=0;i<10;i++)
	{
    
		if(cun[0][i] != 0)
			n1++;
		if(cun[1][i] != 0)
			n2++;
	}
		
//daijia1, Calculate the time sequencing cost of not exchanging cattle  
	int daijiaq=0,daijiah=0;	
	int daijia1=0,daijia2=0;
	
	for(i=0;i<n1;i++)
		daijiaq = daijiaq+cun[0][i];
	daijiaq = daijiaq + cun[0][0]*(n1-2);
	
	for(i=0;i<n2;i++)
		daijiah = daijiah+cun[1][i];
	daijiah = daijiah + cun[1][0]*(n2-2);
	
	daijia1 = daijiaq +daijiah;

//daijia2 , Calculate the time cost of exchanging cattle  , When there is only one stage of rotation , There is no need to consider the exchange problem , take  daijia2 It's very large  

	daijiaq=0;daijiah=0;
	if(cun[1][0]!=0) 
	{
    
		for(i=0;i<n1;i++)
			daijiaq = daijiaq+cun[0][i];		
		for(i=0;i<n2;i++)
			daijiah = daijiah+cun[1][i];
		
		daijiah = daijiah + cun[0][0]*(n2+n1-1)+cun[1][0];	
		daijia2 = daijiaq +daijiah;		
	}
	else
		daijia2 = 9999999;
	
// Deal with head changing  , And put the first result in cun[0][shu] in , Put the second result   Put it in  cun1[i] in  
// When you need to exchange cows  
	if(daijia1>daijia2)
	{
    
		cout<<" The time cost of sorting is ："<<daijia2<<"\n";
		cun1[0] = cun[0][0];
		int i=0;
		for(i=1;i<=n2;i++)
			cun1[i] = cun[1][i-1];
		cun1[n2+1] = cun[1][0];		
	}
	// When there is no need to exchange cows  
	else{
    
		cout<<" The time cost of sorting is ："<<daijia1<<"\n";
		int i=0;
		for(i=0;i<n2;i++)
			cun1[i] = cun[1][i];	
	}
	
// Take again n1 ,n2.
		n1=0,n2=0;
		for(i=0;i<10;i++)
		{
    
			if(cun[0][i] != 0)
				n1++;
			if(cun1[i] != 0)	
				n2++;
		}
		
		
	// Define two variables  diyipai dierpai  To find a cow to exchange with a cow  
	int diyipai = n1,dierpai = n2;
	for(i=0;i<n1-1;i++)
	{
    
		// shou weiba  For the two cows to be exchanged  
		int shou = cun[0][0];
		int weiba =cun[0][diyipai-1];
		diyipai--; 
		// In exchange for  
		int j=0;
		for(j=0;j<n;j++)
		{
    
			if(a[j] == shou)
			{
    
				a[j] = weiba;continue;
			}
			if(a[j] == weiba)
			{
    
				a[j] = shou;continue;
			}
		}
		// Output the exchange results  
		int t=0;
		for(t=0;t<n;t++)
			cout<<a[t]<<" ";
		cout<<"\n";		
	}
	
// Same as the first row above , I'm too lazy to do it anymore ,,（ Brief introduction ~） 
	for(i=0;i<n2-1;i++)
	{
    
		int shou = cun1[0];
		int weiba =cun1[dierpai-1];
		dierpai--; 
		// In exchange for  
		int j=0;
		for(j=0;j<n;j++)
		{
    
			if(a[j] == shou)
			{
    
				a[j] = weiba;	continue;
			}
			if(a[j] == weiba)
			{
    
				a[j] = shou;	continue;
			}
		}
		// Output the exchange results  
		int t=0;
		for(t=0;t<n;t++)
			cout<<a[t]<<" ";
		cout<<"\n";		
	}
}

Um. , It's a long code （ Because I'm stupid ...）

【 Here are the test cases ：】

【 ending 】： Mora , Mora , I hope I can help others （ Manual funny ）, also , Can the boss tell me , Why is it DEV Some test examples on will run incorrectly .