C language · The price of Perfection

Basic exercise perfect cost time limit: 1.0s memory limit: 512.0MBThe brocade SAC 1 uses the greedy algorithm. The SAC 2 enumerates each character from left to right, moving the corresponding character. The number is a single word descriptors the middle. The problem describes a palindrome, which is a special string that reads from left to right and reads from right to left. Longron that palindrome is perfect. Now give you a string, it is not necessarily a palindrome, please calculate the minimum number of exchanges so that the string into a perfect palindrome string.
Interchange is defined as swapping two adjacent characters
such as Mamad
First time Exchange Ad:mamda
Second Exchange MD:MADMA
Third Exchange Ma:madam (palindrome!) Perfect! Input format the first line is an integer n, indicating the length of the next string (n <= 8000)
The second line is a string with a length of n. Only lowercase letter output format if possible, output a minimum number of interchanges.
Otherwise output impossible sample input 5
Mamad Sample Output 3

1 /*2 Perfect Price: The original string is converted to a palindrome string by exchanging adjacent characters. 3 */4#include <stdio.h>5#include <stdlib.h>6 intMain () {7     inti,j,l,n,k,sum=0, flat=1, c=-1;8     Char*A;9scanf"%d",&n);TenA= (Char*)malloc(nsizeof(Char)); Onescanf"%s", a); Aj=n-1;  -     //use the greedy idea to find the number of exchanges required for each character to be traversed and then exchanged to the correct position . -      for(i=0; i<j;i++){ the          for(k=j;k>=i;k--){ -             if(k==i) {//The description did not find the same character as A[i] -                 if(n%2==0|| c!=-1){//if n is even or a[i] is not the only single character without the same characters -flat=0; +                      Break; -                 } +C=1;//n is odd, the number of exchanges required to move the first single character A[i] to an intermediate position Asum=sum+n/2-i; at                  Break; -             } -             if(a[k]==A[i]) { -                  for(l=k;l<j;l++){   -a[l]=a[l+1];  -                 }   ina[j]=A[i];  -sum=sum+j-K;  toj--;  +                  Break;  -             }   the         }   *         if(flat==0){ $              Break;Panax Notoginseng         }  -     }   the     if(flat==0) +printf"Impossible"); A     Else if(sum==0) theprintf"0"); +     Else  -printf"%d\n", sum); $     return 0;  $}

C language · The price of Perfection