ZOJ - Word Reversal(栈)

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1151
Time Limit: 2 Seconds Memory Limit: 65536 KB

Problem Description

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

Problem solving report:

Description: 把每个单词反转，不改变单词的顺序。
Problem solving: 简单的字符串反转，利用栈的性质就可以了。。。

#include 
#include 
#include 
using namespace std;
int main()
{
    char ch; 
    int t1, t2;
    scanf("%d", &t1);
    while (t1--)
    {
        scanf("%d%*c", &t2);
        while (t2--)
        {
            stack  S;
            while (ch = getchar(), ch != '\n')
            {
                if (ch == ' ')
                {
                    while (!S.empty())
                    {
                        printf("%c", S.top());
                        S.pop();
                    }
                    printf(" ");
                }
                else S.push(ch);
            }
            while (!S.empty())
            {
                printf("%c", S.top());
                S.pop();
            }
            printf("\n");    
        }
        if (t1)
            printf("\n");
    }
    return 0;
}