uva 120(排序，检索)

题目：

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.

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

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.

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

题解：翻煎饼。从最小排到最大放好就行。。只能从一个位值将上边的煎饼全部翻了。。找到规律，先将最大的煎饼找到，如果在最上面，就将所有煎饼翻过去，然后找剩下的煎饼里最大的那个，以此类推，如果没找到，就将最大的煎饼翻到最上边。先保存一个排好序的煎饼序列，每次翻好后比对一下，都相同就结束了。。具体看代码。。。

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;
const int M = 40;
int main() {
	int in[M], n = 0, flag, temp[M], temp1[M];
	char c;
	while(scanf("%d%c", &in[n++], &c) != EOF) {
		if (c == '\n') {
			flag = 0;
			for (int i = 0; i < n - 1; i++)
				printf("%d ", in[i]);
			printf("%d\n", in[n - 1]);
			for (int i = 0; i < n; i++) 
				temp[i] = in[i]; sort(in, in + n);
			int k = n, pos = 0, max;
			while (flag = 1) {
				int i, j;
				for (i = 0; i < n; i++)
					if (in[i] != temp[i])
						break;
				if (i == n) {
					flag = 1;
					printf("0\n");
					break;
				}
				max = temp[0], pos = 0;
				for (i = 1; i < k; i++)//先找到最大值的位置
					if (temp[i] > max) {
						max = temp[i];
						pos = i;
					}
				if (pos == k - 1)//如果已经在最后了就让k--最大值变成比他小一个的数
					k--;
				else if (pos == 0) {//如果最大值在开头，就把最大的值放到最后
					printf("%d ", n - k + 1);
					for (i = 0, j = k - 1; i < k, j >= 0; i++, j--)
						temp1[i] = temp[j];
					for (j = 0; j < k; j++) 
						temp[j] = temp1[j];
					k--;
				}
				else {
					printf("%d ", n - pos);
					for (i = pos, j = 0; i >= 0, j <= pos; i--, j++)//把最大的值放到开头
						temp1[j] = temp[i];
					for (j = 0; j <= pos; j++)
						temp[j] = temp1[j];
				}
			}
			n = 0;
		}	 
	}
	return 0;
}