题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problem=46

 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 nwriteln statements must appear in the program.
  • Exactly three semi-colons must appear in the programs
    1. after the program header: program sort(input,output);
    2. after the variable declaration: ...: integer;
    3. 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 n distinct 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.

Miguel Revilla 2001-05-25
 
推荐博客:http://www.cnblogs.com/java20130723/p/3212108.html
 
代码:(没有缩进处理)
 
 #include <cstdio>
const int maxn = ; int n, arr[maxn]; void insert_sort(int p, int c) { //插入排序
for (int i = c; i > p; i--)
arr[i] = arr[i - ];
arr[p] = c;
} int dfs(int d) {
int tmp[d + ]; //创建数组储存原来的数值,不然会乱掉
for (int j = ; j <= n; j++)
tmp[j] = arr[j];
for (int i = d; i >= ; i--) { //循环从现排好的串后序进行dfs
printf("if %c < %c then\n", arr[i] + 'a' - , d + 'a');
insert_sort(i + , d + ); //将接下去的字母插入到i+1的位置
if (d + == n) { //dfs到最深处,输出
printf("writeln(");
printf("%c", arr[] + 'a' - );
for (int j = ; j <= d + ; j++)
printf(",%c", arr[j] + 'a' - );
printf(")\n");
printf("else\n");
}
else {
dfs(d + );
printf("else\n");
}
for (int j = ; j <= n; j++) //还原数组
arr[j] = tmp[j];
}
insert_sort(, d + ); //下面是对最后一个情况,即字母插到整个数组前面,这里是没有else的
if (d + == n) {
printf("writeln(");
printf("%c", arr[] + 'a' - );
for (int j = ; j <= d + ; j++)
printf(",%c", arr[j] + 'a' - );
printf(")\n");
}
else
dfs(d + );
for (int i = ; i <= n; i++)
arr[i] = tmp[i];
} int main() {
int t;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
//前面部分
printf("program sort(input,output);\nvar\n");
printf("a");
for (int i = ; i <= n; i++)
printf(",%c", i + 'a' - );
printf(" : integer;\nbegin\nreadln(");
printf("a");
for (int i = ; i <= n; i++)
printf(",%c", i + 'a' - );
printf(");\n");
dfs(); //开始深搜
printf("end.\n");
if (t != )
printf("\n");
}
return ;
}

Uva 110 - Meta-Loopless Sorts(!循环,回溯!)的更多相关文章

  1. uva 110 Meta-Loopless Sorts 用程序写程序 有点复杂的回溯水题

    题目要求写一个直接用比较排序的pascal程序,挺有趣的一题. 我看题目数据范围就到8,本来以为贪个小便宜,用switch输出. 然后发现比较次数是阶乘级别的,8的阶乘也是挺大的,恐怕会交不上去. 于 ...

  2. UVA 110 Meta-Loopless Sorts(输出挺麻烦的。。。)

     Meta-Loopless Sorts  Background Sorting holds an important place in computer science. Analyzing and ...

  3. uva 387 A Puzzling Problem (回溯)

     A Puzzling Problem  The goal of this problem is to write a program which will take from 1 to 5 puzz ...

  4. UVa 524 Prime Ring Problem【回溯】

    题意:给出n,把从1到n排成一个环,输出相邻两个数的和为素数的序列 照着紫书敲的, 大概就是这个地方需要注意下,初始化的时候a[0]=1,然后dfs(1),从第1个位置开始搜 #include< ...

  5. UVA 140 Brandwidth 带宽 (dfs回溯)

    看到next_permutation好像也能过╮(╯▽╰)╭ 这题学习点: 1.建图做映射 2.通过定序枚举保证字典序最小 3.strtok,sscanf,strchr等函数又复习了一遍,尽管程序中没 ...

  6. UVa 524 Prime Ring Problem(回溯法)

    传送门 Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbe ...

  7. UVA 524 素数环 【dfs/回溯法】

    Description   A ring is composed of n (even number) circles as shown in diagram. Put natural numbers ...

  8. UVA - 12113 Overlapping Squares(dfs+回溯)

    题目: 给定一个4*4的棋盘和棋盘上所呈现出来的纸张边缘,问用不超过6张2*2的纸能否摆出这样的形状. 思路: dfs纸的张数,每一张中枚举这张纸左上角这个点的位置,暴力解题就可以了. 这个题的覆盖太 ...

  9. UVA - 225 Golygons (黄金图形)(回溯)

    题意:平面有k个障碍点.从(0,0)出发,第一次走1个单位,……,第n次走n个单位,恰好回到(0,0),每次必须转弯90°,图形可以自交,但不能经过障碍点.按字典序输出所有移动序列,并输出序列总数. ...

随机推荐

  1. HT图形组件设计之道(一)

    HT for Web简称HT提供了涵盖通用组件.2D拓扑图形组件以及3D引擎的一站式解决方案,正如Hightopo官网所表达的我们希望提供:Everything you need to create ...

  2. java版复利计算器升级

    github地址:https://github.com/iamcarson/Carson 伙伴:彭宏亮 学号:201406114148 与伙伴工作帅照: 本次升级的地方: 1.改善了界面显示,让界面整 ...

  3. C# WinForm 技巧八:界面开发之“WeifenLuo.WinFormsUI.Docking+OutLookBar” 使用

    概述 转自 http://www.cnblogs.com/luomingui/archive/2013/09/19/3329763.html 最近几天一直在关注WinFrom方面的文章 有想着提炼一下 ...

  4. Python基础:数值(布尔型、整型、长整型、浮点型、复数)

    一.概述 Python中的 数值类型(Numeric Types)共有5种:布尔型(bool).整型(int).长整型(long).浮点型(float)和复数(complex). 数值类型支持的主要操 ...

  5. ahjesus Unity3D XML注释被编译的问题

    public class XMLStringReader : MonoBehaviour { public string slectedItem; private bool editing = fal ...

  6. jdk1.8 J.U.C之FutureTask实现机制分析

    我画了一张关于FutureTask的类图,主要包括FutureTask的几个重要的函数和字段,还有它和父类的关系. 根据上面图我们可以清晰的看出FutureTask的继承关系.FutureTask继承 ...

  7. dapper的增、删、查改的CodeSmith模板

    <%@ Template Language="C#" TargetLanguage="Text" %> <%@ Property Name=& ...

  8. [Xamarin.Android] 发布NuGet套件

    [Xamarin.Android] 发布NuGet套件 前言 在Xamarin中,可以将自己开发的项目包装成为NuGet套件发布至NuGet Server,来提供其他开发人员使用.本篇介绍如何封装并发 ...

  9. js填写银行卡号,每隔4位数字加一个空格

    1.原生js写法 !function () { document.getElementById('bankCard').onkeyup = function (event) { var v = thi ...

  10. location对象及history对象

     history对象 location 是最有用的BOM对象之一,它提供了与当前窗口中加载的文档有关的信息,还提供了一些导航功能.事实上,location 对象是很特别的一个对象,因为它既是windo ...