Description

Ignatius likes to write words in reverse way. Given a single line of text which is written by Ignatius, you should reverse all the words and then output them. 
 

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single line with several words. There will be at most 1000 characters in a line. 
 

Output

For each test case, you should output the text which is processed. 
 

Sample Input

3
olleh !dlrow
m'I morf .udh
I ekil .mca
 

Sample Output

 #include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
char a[];
int n;
scanf("%d",&n);
getchar();
while(n--)
{
gets(a);
int len=strlen(a);
int next[],m=;
for(int i=;i<=len;i++)
{
if(a[i]==' '||i==len)
{
for(int j=i-;j>=&&a[j]!=' ';j--)
{
next[m++]=j;
}
}
if(a[i]==' ')
{
next[m++]=i;
}
}
for(int i=;i<len;i++)
{
putchar(a[next[i]]);
}
printf("\n");
}
}
hello world! I'm from hdu. I like acm.

Hint

 Remember to use getchar() to read '\n' after the interger T, then you may use gets() to read a line and process it.
         
 
 

CDZSC_2015寒假新人(1)——基础 g的更多相关文章

  1. CDZSC_2015寒假新人(2)——数学 G

    G - G Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status ...

  2. CDZSC_2015寒假新人(1)——基础 e

    Description Julius Caesar lived in a time of danger and intrigue. The hardest situation Caesar ever ...

  3. CDZSC_2015寒假新人(1)——基础 i

    Description “Point, point, life of student!” This is a ballad(歌谣)well known in colleges, and you mus ...

  4. CDZSC_2015寒假新人(1)——基础 h

    Description Ignatius was born in a leap year, so he want to know when he could hold his birthday par ...

  5. CDZSC_2015寒假新人(1)——基础 f

    Description An inch worm is at the bottom of a well n inches deep. It has enough energy to climb u i ...

  6. CDZSC_2015寒假新人(1)——基础 d

    Description These days, I am thinking about a question, how can I get a problem as easy as A+B? It i ...

  7. CDZSC_2015寒假新人(1)——基础 c

    Description FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the wareho ...

  8. CDZSC_2015寒假新人(1)——基础 b

    Description The highest building in our city has only one elevator. A request list is made up with N ...

  9. CDZSC_2015寒假新人(1)——基础 a

    Description Contest time again! How excited it is to see balloons floating around. But to tell you a ...

随机推荐

  1. C/C++中的sizeof

    代码: #include <iostream> #include <string> using namespace std; int main(){ char s1[]=&qu ...

  2. (原).cc 和 .cpp 后缀结尾的文件的区别

    This caused a few problems the first time C++ was ported to a system where case wasn't significant i ...

  3. Longest Palindromic Substring -LeetCode

    题目 Given a string s,find the longest palindromic substring in S.You may assume  that the maximum len ...

  4. linux for循环

    一定要记得写后面的分号:http://www.runoob.com/linux/linux-shell-variable.html 这个页面的课程的循环教程是有问题的 for color in yel ...

  5. JSONP有什么作用

    1.解决跨域访问数据                 由于同源策略的限制,XmlHttpRequest只允许请求当前源(域名.协议.端口)的资源,为了实现跨域请求,可以通过script标签实现跨域请求 ...

  6. CSS的权重(转)

    CSS写的渐渐多了,他的权重问题就不得不昂首面对,之前一直得过且过的将就用着,直到最近遇到了几个大坑,一直割刺着我对前端的热情,得了得了,蒙不过去了,就发点时间记下来吧,当然还是一片转载的文章,有时候 ...

  7. AngularJS 父子控制器

    <!doctype html> <html ng-app="myApp"> <head> <script src="C:\\Us ...

  8. C语言解析日志,存储数据到伯克利DB

    编译命令 gcc -o dbwriter dbwriter.c -ldb dbwriter.c #include <assert.h> #include <stdlib.h> ...

  9. [POJ] 3461 Oulipo [KMP算法]

    Oulipo Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 23667   Accepted: 9492 Descripti ...

  10. 迁移笔记:对ob_start()的总结

    1.Flush:刷新缓冲区的内容,输出. 函数格式:flush() 说明:这个函数经常使用,效率很高. 2.ob_start :打开输出缓冲区 函数格式:void ob_start(void) 说明: ...