Word Amalgamation

Time Limit: 1 Sec  Memory Limit: 64 MB Submit: 373  Solved: 247

Description

In millions of newspapers across the United States there is a word game called Jumble. The object of this game is to solve a riddle, but in order to find the letters that appear in the answer it is necessary to unscramble four words. Your task is to write a program that can unscramble words.

Input

The input contains four parts: 1) a dictionary, which consists of at least one and at most 100 words, one per line; 2) a line containing XXXXXX, which signals the end of the dictionary; 3) one or more scrambled 'words' that you must unscramble, each on a line by itself; and 4) another line containing XXXXXX, which signals the end of the file. All words, including both dictionary words and scrambled words, consist only of lowercase English letters and will be at least one and at most six characters long. (Note that the sentinel XXXXXX contains uppercase X's.) The dictionary is not necessarily in sorted order, but each word in the dictionary is unique.

Output

For each scrambled word in the input, output an alphabetical list of all dictionary words that can be formed by rearranging the letters in the scrambled word. Each word in this list must appear on a line by itself. If the list is empty (because no dictionary words can be formed), output the line "NOT A VALID WORD" instead. In either case, output a line containing six asterisks to signal the end of the list.

Sample Input

tarp
given
score
refund
only
trap
work
earn
course
pepper
part
XXXXXX
resco
nfudre
aptr
sett
oresuc
XXXXXX

Sample Output

score
******
refund
******
part
tarp
trap
******
NOT A VALID WORD
******
course
******

HINT

 #include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[][] ;
char b[][] ; void Sort (int n)
{
for (int i = ; i < n; i++) {
if (strcmp (a[i] , a[i - ]) < ) {
char ans[] ;
strcpy (ans , a[i] ) ;
int j ;
for (j = i - ; j >= && strcmp (ans , a[j]) < ; j--) {
strcpy (a[j + ] , a[j] ) ;
}
strcpy (a[j + ] , ans ) ;
}
}
} int main ()
{
//freopen ("a.txt" , "r" , stdin ) ;
int n = , s ;
do {
gets (a[n++]) ;
} while (strcmp (a[n - ] , "XXXXXX") != ) ;
n-- ;
Sort (n) ;
int k = ;
do {
gets (b[k++]) ;
} while (strcmp (b[k - ] , "XXXXXX") != ) ;
k-- ;
for (int i = ; i < k ; i++) {
bool flag = ;
for (int j = ; j < n ; j++) {
int l1 , l2 ;
char s1[] ;
char s2[] ;
strcpy (s1 , b[i]) ; l1 = strlen (s1) ; //puts (s1) ;
strcpy (s2 , a[j]) ; l2 = strlen (s2) ; //puts (s2) ;
if (l1 != l2)
continue ;
sort (s1 , s1 + l1 ) ; //puts (s1) ;
sort (s2 , s2 + l2 ) ; //puts (s2) ;
for (s = ; s < l1 ; s++) {
if (s1[s] != s2[s])
break ;
}
if (s == l1) {
printf ("%s\n" , a[j]) ;
flag = ;
}
}
if (!flag) {
puts ("NOT A VALID WORD") ;
puts ("******") ;
}
else {
puts ("******") ;
}
}
return ;
}

Word Amalgamation(枚举 + 排序)的更多相关文章

  1. poj1318 Word Amalgamation 字符串排序(qsort)

    Word Amalgamation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9794   Accepted: 4701 ...

  2. ZOJ1181 Word Amalgamation 字符串 排序查找

    传送门:ZOJ1181  思路:自身排序来判断两个字符串拥有相同的字符.   #include<cstdio> #include<cstdlib> #include<io ...

  3. hdu-----(1113)Word Amalgamation(字符串排序)

    Word Amalgamation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others ...

  4. hdu1113 Word Amalgamation(详解--map和string的运用)

    版权声明:本文为博主原创文章.未经博主同意不得转载. vasttian https://blog.csdn.net/u012860063/article/details/35338617 转载请注明出 ...

  5. HDOJ.1113 Word Amalgamation(map)

    Word Amalgamation 点我挑战题目 点我一起学习STL-MAP 题意分析 给出字典.之后给出一系列======乱序======单词,要求你查字典,如过这个乱序单词对用有多个有序单词可以输 ...

  6. poj 1318 Word Amalgamation

    Word Amalgamation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9968   Accepted: 4774 ...

  7. Uva 642 - Word Amalgamation sort qsort

     Word Amalgamation  In millions of newspapers across the United States there is a word game called J ...

  8. OpenMP之枚举排序

    // EnumSort.cpp : 定义控制台应用程序的入口点. //枚举排序 /* 枚举排序(Enumeration Sort)是一种最简单的排序算法,通常也称为秩排序(Rank Sort). 该算 ...

  9. Word Amalgamation(hdoj1113)

    Word Amalgamation Problem Description In millions of newspapers across the United States there is a ...

随机推荐

  1. Logparser 的用法

    Logparser是一款非常强大的日志分析软件,可以帮助你详细的分析网站日志.是所有数据分析和网站优化人员都应该会的一个软件.Logparser是微软的一款软件完全免费的,大家可以在微软的官网上去下载 ...

  2. EF保存平面数据到SqlServer

    前言 公司开展一个项目,需要根据客户手机定位获取周围内的精准广告,具体是管理员在地图上绘制多边形的广告范围,落在范围内的客户就看到此广告.下面将我的实现方法简单叙述一下,以供有相同需求的朋友参考. E ...

  3. Swift 玩转 3D Touch 之 Peek & Pop

    什么是3D Touch 3D Touch 是iOS9之后专为 iPhone6s 机型加入的新特性,这一新技术移植于 Mac Book 上的 ForceTouch 更准确地说应该是 ForceTouch ...

  4. angular实现了一个简单demo,angular-weibo-favorites

    前面必须说一段 帮客户做了一个过渡期的项目,唯一的要求就是速度,我只是会点儿基础的php,于是就用tp帮客户做了这个项目.最近和客户架构沟通,后期想把项目重新做一下,就用现在最流行的技术,暂时想的使用 ...

  5. Orchard 刨析:Logging

    最近事情比较多,有预研的,有目前正在研发的,都是很需要时间的工作,所以导致这周只写了两篇Orchard系列的文章,这边不能保证后期会很频繁的更新该系列,但我会写完这整个系列,包括后面会把正在研发的东西 ...

  6. SSRS用自定义对象绑定报表

    有一个报表的数据源是一个对象的List, 这个对象List中还有层级,其中还有其他的对象List,这样的层级有三层.其数据是从数据库中取出来的.其LINQ的操作太多了而且复杂,所以不太可 能从LINQ ...

  7. [BZOJ1801][AHOI2009]中国象棋(递推)

    题目:http://www.lydsy.com:808/JudgeOnline/problem.php?id=1801 分析: 只会50的状态压缩…… 然后搜了下题解,发现是dp 首先易得每行每列至多 ...

  8. JavaScript基础---作用域,匿名函数和闭包

    匿名函数就是没有名字的函数,闭包是可访问一个函数作用域里变量的函数. 一.匿名函数 //普通函数 function box() { //函数名是 box return 'TT'; } //匿名函数 f ...

  9. [设计模式] javascript 之 单件模式

    单件模式说明 1. 说明:单件模式,就是静态化的访问中已经实例化的对象,这个对象只能通过一个唯一的入口访问,已经实例或待实例化的对象:面向对象语言如Java, .Net C#这样的服务端动态语言里,能 ...

  10. Moqui学习之数据与资源

    资源位置: 资源门面位置的字符串类似于URL的构成方式:协议,主机,可选端口和文件名.它支持标准的java URL协议(http https ftp jar file).同样也支持一些扩展的协议: c ...