Uva 110 - Meta-Loopless Sorts(!循环,回溯!)
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 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
- 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 ...
- 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 ...
- 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 ...
- 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< ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- Java 集合系列02之 Collection架构
概要 首先,我们对Collection进行说明.下面先看看Collection的一些框架类的关系图: Collection是一个接口,它主要的两个分支是:List 和 Set. List和Set都是接 ...
- Android开发总结
出来工作半年多了,没啥好交代的,就说说自己半年来的Android开发经历. 1.IDE 这半年来,从Eclipse到Android Studio,经历了两个IDE,在这里做一下简单的评价. ...
- C#对象与XMl文件之间的相互转换
C#提供三种序列化方式,分别为: 1.是使用BinaryFormatter进行串行化: 2.使用SoapFormatter进行串行化: 3.使用XmlSerializer进行串行化.其中对于Binar ...
- EF错误记录
纯属个人记录错误使用: 1.EntityType“area”未定义键.请为该 EntityType 定义键. 产生原因: 1.命名空间引用错误,可能命名重复导致引用错误 2.实体类无法识别主键或者未设 ...
- Struts2的基本流程的详细介绍
Struts2基本流程 概述: Struts2框架由三部分构成:核心控制器.业务控制器和用户实现的业务逻辑组件.在这三部分中,struts2框架提供了核心控制器StrutsPrepareAndExec ...
- Python入门笔记(9):元组
一.元组特性 1.类似列表,但不可变类型,正因如此,它可以做一个字典的key2.当处理一组对象时,这个组默认是元组类型(老写错"元祖")3.所有的多对象,逗号分隔的,没有明确用符号 ...
- Free Slideshow, Gallery And Lightboxes Scripts
http://bootstraphelpers.codeplex.com/SourceControl/list/changesets https://github.com/gordon-matt/Bo ...
- ActiveReports 报表应用教程 (7)---交叉报表及数据透视图实现方案
在 ActiveReports 中可以通过矩阵控件非常方便的实现交叉报表,同时还可以设置数据的分组.排序.过滤.小计.合计等操作,可以满足您报表的智能数据分析等需求.在矩阵控件中组的行数和列数由每个行 ...
- ACdream 1214---矩阵连乘
ACdream 1214---矩阵连乘 Problem Description You might have noticed that there is the new fashion among r ...
- [Tool] 使用Visual Studio Code开发TypeScript
[Tool] 使用Visual Studio Code开发TypeScript 注意 依照本篇操作步骤实作,就可以在「Windows」.「OS X」操作系统上,使用Visual Studio Code ...