zoj 1201

Inversion

 Let { A1,A2,...,An } be a permutation of the set{ 1,2,..., n}. If i < j and Ai > Aj then the pair (Ai,Aj) is called an "inversion" of the permutation. For example, the permutation {3, 1, 4, 2} has three inversions: (3,1), (3,2) and (4,2).

  The inversion table B1,B2,...,Bn of the permutation { A1,A2,...,An } is obtained by letting Bj be the number of elements to the left of j that are greater than j. (In other words, Bj is the number of inversions whose second component is j.) For example, the permutation:
{ 5,9,1,8,2,6,4,7,3 }
has the inversion table
2 3 6 4 0 2 2 1 0
since there are 2 numbers, 5 and 9, to the left of 1; 3 numbers, 5, 9 and 8, to the left of 2; etc.
  Perhaps the most important fact about inversions is Marshall Hall's observation that an inversion table uniquely determines the corresponding permutation. So your task is to convert a permutation to its inversion table, or vise versa, to convert from an inversion table to the corresponding permutation.

Input:
The input consists of several test cases. Each test case contains two lines.
The first line contains a single integer N ( 1 <= N <= 50) which indicates the number of elements in the permutation/invertion table. 
The second line begins with a single charactor either 'P', meaning that the next N integers form a permutation, or 'I', meaning that the next N integers form an inversion table. 
Following are N integers, separated by spaces. The input is terminated by a line contains N=0.

Output:
For each case of the input output a line of intergers, seperated by a single space (no space at the end of the line). If the input is a permutation, your output will be the corresponding inversion table; if the input is an inversion table, your output will be the corresponding permutation.

Sample Input:

9
P 5 9 1 8 2 6 4 7 3
9
I 2 3 6 4 0 2 2 1 0
0

Sample Output:

2 3 6 4 0 2 2 1 0
5 9 1 8 2 6 4 7 3

代码：

#include <stdio.h>
#include <string.h>
#include<malloc.h>


typedef struct Lt
{
	int m;
    struct	Lt *next;
} Lt,*LP;

int main()
{
	int n,i;
	int input[52];
	char ch;
	
	void Invertion (int input[],int n);
    void Permutation(int input[],int n);
    
	while(scanf("%d",&n)!=EOF&&n!=0)
	{
		getchar();
		scanf("%c",&ch);
		memset(input,0,sizeof(input));
		for (i=1;i<=n;i++)
		{
			scanf("%d",&input[i]);
			
		} 

		switch(ch)
		{
		case  'P':
		    Invertion (input,n);
		    break;
	    case  'I':
		    Permutation(input,n);
			break;
		default:
	    	return 0;
		}

	}

	return 0;
}

void Invertion (int input[],int n)
{
	int i,j;
	int *m=malloc((n+1)*sizeof(int));
	for(i=0;i<=n;i++)
	{
		m[i]=0;
	}
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=i;j++)
		{
			if (input[j]>input[i])
			{
				m[input[i]]++;
				
			}
		}
	}
	for(i=1;i<n;i++)
	{
		printf("%d ",m[i]);
	}
	printf("%d\n",m[n]);
	free(m);
}

void Permutation(int input[],int n)
{
	

	int i,j;
	LP p,q,s;

    LP slist=(LP)malloc(sizeof(Lt)); 
	slist->m=0;
	slist->next=NULL;

	for(i=n;i>=1;i--)
	{
	 
		q=(LP)malloc(sizeof(Lt));
		q->m=i;
		q->next=NULL;

		j=0;
		p=slist;

		while(j<input[i])
		{
			p=p->next;
			j++;
		}
		s=p->next;
		p->next=q;
	    q->next=s;
			
	}

	for(p=slist->next;p->next!=NULL;p=p->next)
	{
		printf("%d ",p->m);
	}
	printf("%d\n",p->m);

	while (slist->next!=NULL)
	{
	
		q=slist->next;
		slist->next=q->next;
		free(q);

	}
	free(slist);


}