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. iOS端给unity发送消息,实现两者交互。

    上一篇我们简单说了一下unity发消息给iOS端.现在我们就来说一下iOS端给unity发送消息的简单使用. 首先iOS端做得事情其实很简单就一句话,直接上代码 /** * 第一个参数:是unity那 ...

  2. Windows Phone8 中如何引用 SQLite 数据库

    SQLite数据库介绍 1.SQLite是一款轻型的嵌入式数据库,使用C++开发,使用非常广泛 2.SQLite是一款跨平台的数据库,支持Windows.Linux.Android.IOS.Windo ...

  3. State Pattern -- 状态模式原理及实现(C++)

    主要参考<大话设计模式>和<设计模式:可复用面向对象软件的基础>两本书.本文介绍命令模式的实现. 问题出发点 在实际开发中,我们经常会遇到这种情况:一个对象有多种状态,在每一个 ...

  4. 第四章 一切从IL开始

    从这一部分开始,就开始讲.net的本质了,这第四章就是讲有关IL(中间语言)的内容,主要利用工具,看看VS到底编译出来的是什么东西,从中我们可以看到实现的方式和过程.有助于我们更好的了解.net的本质 ...

  5. Android图片浏览器之缩略图

    项目源码:http://files.cnblogs.com/files/tgyf/app.rar. 最近在自学Android,尝试实现一般手机上都存在的图片浏览器,从缩略图开始. 直接上图,这是goo ...

  6. SequoiaDB 系列之二 :SequoiaDB的简单CRUD操作

    上一篇通过一系列的操作,终于把SequoiaDB的集群部署到单台机器上了. 建议去安装体验一下吧. 在整个环境的部署的体验来看,并没有MongoDB的部署简单,但是比MongoDB的部署要清晰.Mon ...

  7. javaee添加验证码

    package com.zh.code; import java.awt.Color; import java.awt.Font; import java.awt.Graphics; import j ...

  8. winform之权限判断小技巧

    每个页面都要判断用户是否登陆并且判断用户是否拥有相应的权限,,以至于每个页面都要判断Session["user"]是否为空,后期不好维护 小技巧: 因为每个页面都继承与Page类, ...

  9. Oracle自定义函数实例

    1. 传入一个值, 如果该值为0,则返回空. CREATE OR REPLACE FUNCTION Fun_Test(p IN NUMBER) RETURN VARCHAR2 IS v_Result ...

  10. ethtool使用记录

    网卡出现很诡异的问题,把电脑连到一些交换机上是工作的,连到另外一些就不行...交换机上的link灯还时不时的闪一下,看起来像是在尝试连接. 用dmesg查看,看到下面的信息: [ 1112.92211 ...