ZOJ Problem Set - 1151 Word Reversal

Word Reversal

Time Limit: 2 Seconds Memory Limit: 65536 KB

For each list of words, output a line with each word reversed without changing the order of the words.

This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank
line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.

Input
You will be given a number of test cases. The first line contains a positive integer indicating the number of cases to follow. Each case is given on a line containing a list of words
separated by one space, and each word contains only uppercase and lowercase letters.

Output
For each test case, print the output on one line.

Sample Input
1
3

I am happy today

To be or not to be

I want to win the practice contest

Sample Output
I ma yppah yadot

oT eb ro ton ot eb

I tnaw ot niw eht ecitcarp tsetnoc

Source: East Central North America 1999, Practice

分析：

题目很简单，坑就坑在字符串的处理上。而且感觉题意也不清楚。但无论如何作为一个加强对字符串处理理解的问题，还是有一定价值的。

题意：

给定若干字符串（分块），输出它们的反串（反过来）。

ac代码：

//scanf()是无法判断读入的是不是空格，换行之类符号的

//这里应该是认为单词之间必定有空格，即没有一行只有一个单词的情况

#include <iostream>

#include<cstdio>

#include<cstring>

#include<string>

using namespace std;

char c[100];

string s,ss;

int main()

{

int n,m,i;

cin.getline(c,5);//不处理块数量，这里同时也把下一行的换行读取了

cin.getline(c,5);//不处理第一块的行数量

bool flag=1;

while(cin.getline(c,100))

{

s=c;

//int L=s.size();

if(s.find(' ')==2147483647*2+1&&!flag)//是行数量又不是第一块.这里应该是认为单词之间必定有空格，即没有一行只有一个单词的情况

{

cin.getline(c,5);

cout<<endl;

}

else flag=0;

if(s.find(' ')!=2147483647*2+1)//这里应该是认为单词之间必定有空格，即没有一行只有一个单词的情况

{

int L=s.size();

for(i=0;i<L;i++)

{

if(s[i]!=' ')

ss=s[i]+ss;

else

{

cout<<ss<<" ";

ss="";

}

}

cout<<ss<<endl;

ss="";

}

}

return 0;

}