#|ZOJ1151 Word Reversal【堆栈+字符流+字符串流】 #|ICPC-数据结构|#|ICPC-ZOJ|#|ICPC文

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
【#|ZOJ1151 Word Reversal【堆栈+字符流+字符串流】】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
问题链接：ZOJ1151 Word Reversal
问题简述：（略）
问题分析：
这是一个简单的文本处理问题。用C++实现的话，可以使用字符串留来处理。正道还是用字符流和堆栈来解决。
程序说明：（略）
参考链接：（略）
题记：（略）
AC的C语言程序如下：

/* ZOJ1151 Word Reversal */#include #define MAXSTACK 1024char stack[MAXSTACK]; int pstack; void push(char c) { stack[pstack++] = c; }char pop() { return stack[--pstack]; }int main(void) { int t, line, i; char c; scanf("%d", &t); while(t--) { scanf("%d", &line); getchar(); pstack = 0; for(i=1; i<=line; i++) { c = getchar(); while(c != '\n') { if(c == ' ') { while(pstack) putchar(pop()); putchar(c); } else push(c); c = getchar(); }while(pstack) putchar(pop()); putchar('\n'); }if(t) putchar('\n'); }return 0; }

AC的C++语言程序如下：

/* ZOJ1151 Word Reversal */#include #include #include using namespace std; void reverse(string& s) { string t; stringstream ss(s); bool flag = false; while(ss >> t) { if(flag) cout << ' '; flag = true; for(int i = (int)t.size() - 1; i >= 0; i--) cout << t[i]; } cout << endl; }int main() { int t, n; cin >> t; while(t--) { cin >> n; cin.get(); string s; for(int i = 1; i <= n; i++) { getline(cin, s); reverse(s); } if(t) cout << endl; }return 0; }