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ZOJ|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




【ZOJ|ZOJ Problem Set - 1151 Word Reversal】

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
#include
#include
#include
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<}
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{
if(s[i]!=' ')
ss=s[i]+ss;
else
{
cout}
}
cout}
}
return 0;
}



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