#|ZOJ - Word Reversal(栈) ACM题解|数据结构

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!
【#|ZOJ - Word Reversal(栈)】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; }