uva 120 Stacks of Flap jacks

题意 翻转烧饼 每次翻转是从自选中的烧饼往上到顶部一整叠一起翻转 最终要得到从小到大排列的烧饼

解法 把烧饼从大到小每一个都进行一次判断，如果不在该放置的位置，就把这烧饼翻到顶，再翻到底，直到每个烧饼都判断完成。

代码:

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
#include <algorithm>

int stack[105], t;
int num[105];
char c;
void tra(int k)
{
    for (int i = 0; i < k/2; i ++)
    {
	int sb = stack[i];
	stack[i] = stack[k - 1 - i];
	stack[k - 1 - i] = sb;	
    }	
}
int findmax(int k)
{
    int max;
    max = stack[k - 1];
    for (int i = k - 1; i >= 0; i --)
    {
	if (stack[i] > max)
	    max = stack[i];
    }
    for (int i = k - 1; i >= 0; i --)
    {
	if (max == stack[i])
	    return i;
    }
}
int judge(int t)
{
    for (int i = 0; i < t - 1; i ++)
    {
	if(stack[i] > stack[i + 1])
	    return 1;
    }
    return 0;
}
int main()
{   
    t = 0;
    memset(stack, 0, sizeof(stack));
    while (scanf("%d%c", &stack[t], &c) != EOF)
    {
	t ++;
	if (c == '\n')
	{
	    for (int i = 0; i < t - 1; i ++)
	    {
		printf("%d ", stack[i]);
	    }
	    printf("%d\n", stack[t - 1]);
	    int j = 0;
	    int k = t;
	    while (judge(t))
	    {
		if(findmax(k) == 0 && findmax(k) != k - 1)
		{
		    num[j++] = t - k + 1;
		    tra(k);
    
		}
		if(findmax(k) != 0 && findmax(k) != k - 1)
		{
		    num[j++] = t - findmax(k);
		    tra(findmax(k) + 1);
		    num[j++] = t - k + 1;
		    tra(k);
		}
		k --;
	    }
	    num[j] = 0;
	    for(int i = 0; i < j; i ++)
	    {
		printf("%d ",num[i]);
	    }
	    printf("%d\n", num[j]);
	    t = 0;
	    memset(num, 0, sizeof(num));
	    memset(stack, 0, sizeof(stack));
	}
    }
    return 0;
}