ZOJ-1151 Word Reversal，栈的解法

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

******栈法求解******************************************************************************************************************

#include<stdio.h>  
#include<algorithm>  
#include<iostream>  
#include<stack>  
#include<string.h>  
using namespace std;  
stack<char>z;  
int main()  
{  
    int k;  
    int n;  
    char a[999];  
    scanf("%d",&k);  
    while(k--)  
    {  
        scanf("%d",&n);  
        getchar();  
        while(n--)  
        {  
            gets(a);  
            int l=strlen(a);  
            int i=0;  
            while(1)  
            {  
                while(a[i]!=' '&&i<l)  
                {  
                    z.push(a[i]);  
                    i++;  
                }  
                while(!z.empty())  
                {  
                    printf("%c",z.top());  
                    z.pop();  
                }  
                if(i!=l)  
                {  
                    printf(" ");  
                    i++;  
                }  
                else break;  
            }  
            printf("\n");  
        }  
        if(k)  
            printf("\n");  
    }  
    return 0;  
}