uva 110 Meta-Loopless Sorts
题目地址：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=46
题目描述：
| Meta-Loopless Sorts |
Background
Sorting holds an important place in computer science. Analyzing and implementing various sorting algorithms forms an important part of the education of most computer scientists, and sorting accounts for a significant percentage of the world's computational resources. Sorting algorithms range from the bewilderingly popular Bubble sort, to Quicksort, to parallel sorting algorithms and sorting networks. In this problem you will be writing a program that creates a sorting program (a meta-sorter).The Problem
The problem is to create several programs whose output is a standard Pascal program that sorts n numbers where n is the only input to the program you will write. The Pascal programs generated by your program must have the following properties:- They must begin with program sort(input,output);
- They must declare storage for exactly n integer variables. The names of the variables must come from the first n letters of the alphabet (a,b,c,d,e,f).
- A single readln statement must read in values for all the integer variables.
- Other than writeln statements, the only statements in the program are if then else statements. The boolean conditional for each if statement must consist of one strict inequality (either < or >) of two integer variables. Exactly n! writeln statements must appear in the program.
- Exactly three semi-colons must appear in the programs
- after the program header: program sort(input,output);
- after the variable declaration: ...: integer;
- after the readln statement: readln(...);
- No redundant comparisons of integer variables should be made. For example, during program execution, once it is determined that a < b, variables a and b should not be compared again.
- Every writeln statement must appear on a line by itself.
- The programs must compile. Executing the program with input consisting of any arrangement of any ndistinct integer values should result in the input values being printed in sorted order.
For those unfamiliar with Pascal syntax, the example at the end of this problem completely defines the small subset of Pascal needed.
The Input
The input consist on a number in the first line indicating the number M of programs to make, followed by a blank line. Then there are M test cases, each one consisting on a single integer n on a line by itself with 1 n 8 . There will be a blank line between test cases.The Output
The output is M compilable standard Pascal programs meeting the criteria specified above. Print a blank line between two consecutive programs.Sample Input
1 3
Sample Output
program sort(input,output);
var
a,b,c : integer;
begin
readln(a,b,c);
if a < b then
if b < c then
writeln(a,b,c)
else if a < c then
writeln(a,c,b)
else
writeln(c,a,b)
else
if a < c then
writeln(b,a,c)
else if b < c then
writeln(b,c,a)
else
writeln(c,b,a)
end.
题意：写一个如上述 输出形式的排序程序，并且这里面还包含了 排序的结果。
题解：一开始想到用打表 毕竟只有8个字母要排序，后来想想不用打表 也可以很简单地搞出来。
我是用插空的思想来做的： 考虑只有1个字母 a 那么 用字母a去 插空 一个空集 及只有1个空可以插 那么 答案是 a
考虑有 2个字母 a，b 那么 a只有 一种插法 在递归到b 用b去插剩余的集合（只有a），那么有两个空可以插 （_a_）那么 答案是 ab ,ba
考虑三个字母 a，b，c 这里 由于前面已经讨论好a和b的插法了 直接看c 的插空 就是 （_a_b_）和 （_b_a_） 这样可以得到6个答案。
。。。。。。后面以此类推
如何将刚才的分析对应于程序的书写呢， 在此用3个字母的做样例说明：
首选考虑a的插法 程序中是没有体现的 那么直接考虑 b的插空 程序中的体现是 if a < b then 和 对应的 else 而这两条语句分别对应的是b插在a后面 和 b插在a前面 两个动作
然后 看下一级的c 以此类推 内推 分别是 c插 最后面 ， c插 中间 和 c插在最前面。
以上：
按照这样的插空规律可以写出一个递归的程序。
代码：
/*
topcoder codeforce acm
*/
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string>
#include<vector>
using namespace std;
int T=0,N=0;
char ch[10]={'a','a','b','c','d','e','f','g','h','i'};
vector<int> List;
/*print the 2*n ' '*/
int PrintIndent(int n)
{
int m=n+n;
int i=0;
for(i=0;i<=m-1;i++)
{
printf(" ");
}
return(0);
}
/*if else recursive,the n is the index of the ch not the List*/
int IfElse(char c,int n)
{
if(n>N)
{
PrintIndent(n-1);
printf("writeln(");
int i=0;
printf("%c",List[0]);
for(i=1;i<=N-1;i++)
{
printf(",%c",List[i]);
}
printf(")\n");
}
else
{
int i=List.size()-1;//the pointer to the end of the vector
//if structure
PrintIndent(n-1);
printf("if ");
printf("%c < %c",List[i],c);
printf(" then\n");
//if's action
List.insert(List.begin()+i+1,c);//insert the empty blank
IfElse(ch[n+1],n+1);
List.erase(List.begin()+i+1);//because of the recursive we should recover for the present level's state
i--;
//else if structure
for(;i>=0;i--)
{
PrintIndent(n-1);
printf("else if ");
printf("%c < %c",List[i],c);
printf(" then\n");
//else if action
List.insert(List.begin()+i+1,c);
IfElse(ch[n+1],n+1);
List.erase(List.begin()+i+1);
}
//else structure
i=0;
PrintIndent(n-1);
printf("else\n");
//else action
List.insert(List.begin()+i,c);
IfElse(ch[n+1],n+1);
List.erase(List.begin()+i);
}
return(0);
}
/*for test*/
int test()
{
return(0);
}
/*main process*/
int MainProc()
{
scanf("%d",&T);
while(T--)
{
//intialize the vector
List.clear();
List.push_back('a');
scanf("%d",&N);
printf("program sort(input,output);\nvar\na");
int i=0;
for(i=2;i<=N;i++)
{
printf(",%c",ch[i]);
}
printf(" : integer;\n");
//the start of the program
printf("begin\n");
PrintIndent(1);
printf("readln(a");
for(i=2;i<=N;i++)
{
printf(",%c",ch[i]);
}
printf(");\n");
//recursive if else structure
IfElse(ch[2],2);
//end of the program
printf("end.\n");
if(T>0)
{
printf("\n");
}
}
return(0);
}
int main()
{
MainProc();
return(0);
}