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 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.
题意:就是模拟pascal用if,else进行排序。。输入n。输出n个字母的排序过程
思路:大概就是插入排序。一个字母一个字母插进去。每个字母都可以插到当前串的各个位置中去。。。进行回溯输出即可。。但是这输出格式真的挺麻烦的啊啊啊
#include <stdio.h>
#include <string.h> int t, n, i;
int cc[10];
void swap(int &a, int &b)
{
int t = a;
a = b;
b = t;
}
void insert(int num, int *cc) {
int i;
if (num == n)
{
int nu = num * 2;
while (nu --) {printf(" "); }
printf("writeln(");
for (i = 0; i < n - 1; i++) {
printf("%c,", cc[i] + 'a');
}
printf("%c)\n", cc[n - 1] + 'a');
return;
}
int c[10];
for (i = 0; i < num; i ++)
c[i] = cc[i];
for (i = num; i >= 0; i --) {
if (num == i) {
int nu = num * 2;
while (nu --) {printf(" "); }
c[i] = i;
printf("if %c < %c then\n", c[i - 1] + 'a', num + 'a');
insert(num + 1, c);
}
else if (i == 0) {
int nu = num * 2;
while (nu --) {printf(" "); }
printf("else\n");
swap(c[i], c[i + 1]);
insert(num + 1, c);
}
else {
int nu = num * 2;
while (nu --) {printf(" "); }
printf("else if %c < %c then\n", c[i - 1] + 'a', num + 'a');
swap(c[i], c[i + 1]);
insert(num + 1, c);
}
}
}
void out() {
printf("program sort(input,output);\n");
printf("var\n");
for (i = 0; i < n - 1; i++) {
printf("%c,",'a' + i);
}
printf("%c : integer;\n", 'a' + n - 1);
printf("begin\n");
printf(" readln(");
for (i = 0; i < n - 1; i++) {
printf("%c,",'a' + i);
}
printf("%c);\n", 'a' + n - 1);
insert(1, cc);
printf("end.\n");
if (t) printf("\n");
} int main() {
scanf("%d", &t);
while (t --) {
memset(cc, 0, sizeof(cc));
scanf("%d", &n);
out();
}
return 0;
}
UVA 110 Meta-Loopless Sorts(输出挺麻烦的。。。)的更多相关文章
- Uva 110 - Meta-Loopless Sorts(!循环,回溯!)
题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&pa ...
- 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 11419SAM I AM(输出 最小覆盖点 )
参考博客:如何找取 最小覆盖点集合 题意:R*C大小的网格,网格上面放了一些目标.可以再网格外发射子弹,子弹会沿着垂直或者水平方向飞行,并且打掉飞行路径上的所有目标,计算最小多少子弹,各从哪些位置发射 ...
- Euler Circuit UVA - 10735(混合图输出路径)
就是求混合图是否存在欧拉回路 如果存在则输出一组路径 (我就说嘛 咱的代码怎么可能错.....最后的输出格式竟然w了一天 我都没发现) 解析: 对于无向边定向建边放到网络流图中add(u, v, 1) ...
- Hard Life UVA - 1389(最大密度子图 输出点集)
题意: rt 解析: 我用的第二种方法... s向所有的边连权值为1的边 所有的点向t连权值为mid的边 如果存在u - > v 则边向u和v分别连一条权值为INF的边 二分mid 用dfs ...
- uva 400 Unix ls 文件输出排版 排序题
这题的需要注意的地方就是计算行数与列数,以及输出的控制. 题目要求每一列都要有能够容纳最长文件名的空间,两列之间要留两个空格,每一行不能超过60. 简单计算下即可. 输出时我用循环输出空格来解决对齐的 ...
- SAM I AM UVA - 11419(最小顶点覆盖+输出一组解)
就是棋盘问题输出一组解 https://blog.csdn.net/llx523113241/article/details/47759745 http://www.matrix67.com/blog ...
- decode-string(挺麻烦的)
Java String作为参数传参是不会改变的,这个与常识的感觉不同. public String decodeString(String s) { s = ""; return ...
- UVA题解二
UVA题解二 UVA 110 题目描述:输出一个Pascal程序,该程序能读入不多于\(8\)个数,并输出从小到大排好序后的数.注意:该程序只能用读入语句,输出语句,if语句. solution 模仿 ...
随机推荐
- PHP unlink() 函数
定义和用法 unlink() 函数删除文件. 若成功,则返回 true,失败则返回 false. 语法 unlink(filename,context) 参数 描述 filename 必需.规定要删除 ...
- 读写UTF-8、Unicode文件(加上了文件头,貌似挺好用)
conf配置文件一些为UTF-8和Unicode格式,这样便可良好的支持多语言,从网上查阅资料后,将读写UTF-8.Unicode文件写了几个最精简的函数,更新后加了是否写文件头的功能,以适应更多需要 ...
- 二层安全之MAC Flooding解析与解决方法
一.了解MAC Flooding原理 1.1 如图所示,网络中有3个PC和一个交换机,在正常情况下,如果PC A向PC B发送信息,PC C是不会知道的,过程都通过中间的交换机进行透明的处理,并且会记 ...
- JAVA bean与XML互转的利器---XStream
最近在项目中遇到了JAVA bean 和XML互转的需求, 本来准备循规蹈矩使用dom4j忽然想起来之前曾接触过的XStream, 一番研究豁然开朗,利器啊利器, 下来就XStream的一些用法与大家 ...
- 在eclipse里卸载已安装的插件[例如Android Development Tools ADT]
在eclipse里卸载已安装的插件 有四种方法: 1.到plugins和features目录中找到你要卸载的插件的文件夹, ...
- visual studio 2012更换皮肤、功能添加
首先在vs2012的菜单:工具->扩展和更新,打开扩展和更新窗口,点击左侧“联机”,搜索栏里面输入Theme Editor.然后点击按钮,安装之后,在工具->选项->环境常规 面板上 ...
- -_-#【Cookie】缩小 Cookie
Reduce Cookie Size Cookie 是个很有趣的话题.根据 RFC 2109 的描述,每个客户端最多保持 300 个 Cookie,针对每个域名最多 20 个 Cookie (实际上多 ...
- Azure 媒体服务支持 DASH 实时传送流
Kilroy Hughes Azure媒体服务数字媒体架构师 本文重点介绍 Azure 媒体服务支持的 DASH 实时传送流功能,同时阐述如何利用这些功能将实时和点播自适应流传送至 Web 浏览器 ...
- 【HtmlParser】HtmlParser使用
转载 http://www.cnblogs.com/549294286/archive/2012/09/04/2670601.html HTMLParser的核心模块是org.htmlparser.P ...
- 布隆过滤器(Bloom Filter)的原理和实现
什么情况下需要布隆过滤器? 先来看几个比较常见的例子 字处理软件中,需要检查一个英语单词是否拼写正确 在 FBI,一个嫌疑人的名字是否已经在嫌疑名单上 在网络爬虫里,一个网址是否被访问过 yahoo, ...