说明：这道题初读有点绕....开始我以为只是将一个序列的最大值放到最右边，把最小值放到最左边...所以就怎么也通不过啦。后来终于理解题意了，其实就是要你排个序，从左到右为从大到小。但是改变序列顺序的规则有些特殊，即只能将从最左边开始到某一位置构成的连续子序列进行对称翻转。如有序列:  2,4,9,5,7,6.我选择的位置是5所在的位置，翻转后序列就变为5,9,4,2,7,6.然后用类似于冒泡排序的方法。通过最多两次翻转可以得到最大值，然后放到右边。不断重复上述操作，即可得到一个有序序列。

题目：

 Stacks of Flapjacks

Background

Stacks and Queues are often considered the bread and butter of data structures and find use in architecture, parsing, operating systems, and discrete event simulation. Stacks are also important in the theory of formal languages.

This problem involves both butter and sustenance in the form of pancakes rather than bread in addition to a finicky server who flips pancakes according to a unique, but complete set of rules.

The Problem

Given a stack of pancakes, you are to write a program that indicates how the stack can be sorted so that the largest pancake is on the bottom and the smallest pancake is on the top. The size of a pancake is given by the pancake‘s diameter. All pancakes in
a stack have different diameters.

Sorting a stack is done by a sequence of pancake ``flips‘‘. A flip consists of inserting a spatula between two pancakes in a stack and flipping (reversing) the pancakes on the spatula (reversing the sub-stack). A flip is specified by giving the position
of the pancake on the bottom of the sub-stack to be flipped (relative to the whole stack). The pancake on the bottom of the whole stack has position 1 and the pancake on the top of a stack of
n pancakes has position n.

A stack is specified by giving the diameter of each pancake in the stack in the order in which the pancakes appear.

For example, consider the three stacks of pancakes below (in which pancake 8 is the top-most pancake of the left stack):

         8           7           2
         4           6           5
         6           4           8
         7           8           4
         5           5           6
         2           2           7

The stack on the left can be transformed to the stack in the middle via flip(3). The middle stack can be transformed into the right stack via the command
flip(1).

The Input

The input consists of a sequence of stacks of pancakes. Each stack will consist of between 1 and 30 pancakes and each pancake will have an integer diameter between 1 and 100. The input is terminated by end-of-file. Each stack is given as a single line of
input with the top pancake on a stack appearing first on a line, the bottom pancake appearing last, and all pancakes separated by a space.

The Output

For each stack of pancakes, the output should echo the original stack on one line, followed by some sequence of flips that results in the stack of pancakes being sorted so that the largest diameter pancake is on the bottom and the smallest on top. For each
stack the sequence of flips should be terminated by a 0 (indicating no more flips necessary). Once a stack is sorted, no more flips should be made.

Sample Input

1 2 3 4 5
5 4 3 2 1
5 1 2 3 4

Sample Output

1 2 3 4 5
0
5 4 3 2 1
1 0
5 1 2 3 4
1 2 0

源代码：

#include <stdio.h>
#define MAXN 30+5

int pancakes[MAXN];

int main(){
 int i,j,k,cnt=0,max,pos,temp;
 char c;
// freopen("data","r",stdin);
 while(scanf("%d%c",&pancakes[cnt++],&c)==2){
  if(c=='\n'){

   for(i=0;i<cnt;i++)
    printf("%d%c",pancakes[i],i==cnt-1?'\n':' ');

   for(i=1;i<cnt;i++){
    max=pancakes[0];
    pos=0;
    for(j=1;j<=cnt-i;j++){
     if(pancakes[j]>max){
      max=pancakes[j];
      pos=j;
     }
    }
    if(pos==cnt-i)//若最大值已在bottom则无需flip
     continue;

    if(pos!=0){//先将最大值翻转到top
    for(j=0;j<=pos/2;j++){
     temp=pancakes[j];
     pancakes[j]=pancakes[pos-j];
     pancakes[pos-j]=temp;
    }
    printf("%d ",cnt-pos);
    }

    for(j=0;j<=(cnt-i)/2;j++){//再将最大值翻转到bottom
     temp=pancakes[j];
     pancakes[j]=pancakes[cnt-i-j];
     pancakes[cnt-i-j]=temp;
    }
    printf("%d ",i);
   }
   cnt=0;
   printf("0\n");
  }
 }
 return 0;
}

Stacks of Flapjacks UVA 120