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hdu 1062 Text Reverse
Text Reverse |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 6265 Accepted Submission(s): 1713 |
Problem Description Ignatius likes to write words in reverse way. Given a single line of text which is written by Ignatius, you should reverse all the words and then output them. |
Input The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case contains a single line with several words. There will be at most 1000 characters in a line. |
Output For each test case, you should output the text which is processed. |
Sample Input 3olleh !dlrowm‘I morf .udhI ekil .mca |
Sample Output hello world!I‘m from hdu.I like acm. Hint Remember to use getchar() to read ‘\n‘ after the interger T, then you may use gets() to read a line and process it. |
水题，注意细节。
PE三次的代码：
1 #include<cstdio> 2 #include<cstring> 3 #include<stack> 4 using namespace std; 5 const int maxn=1000+10; 6 char str[maxn]; 7 int main() 8 { 9 stack<char>s;10 int T;11 char ch;12 scanf("%d",&T);13 getchar();14 while(T--)15 {16 //s.clear;17 gets(str);18 //getchar();19 for(int i=0;i<strlen(str);i++)20 {21 if(str[i]!=‘ ‘)22 {23 s.push(str[i]);24 }25 if(str[i]==‘ ‘||i==(strlen(str)-1))26 {27 while(!s.empty())28 {29 ch=s.top();30 printf("%c",ch);31 s.pop();32 }33 if(i!=(strlen(str)-1))34 printf(" ");35 }36 }37 printf("\n");38 }39 }
AC的代码：
#include<stdio.h>#include<stack>using namespace std;int main(){ int n; char ch; scanf("%d",&n); getchar(); /*吸收回车符*/ while(n--) { stack<char> s; /*定义栈*/ while(true) { ch=getchar(); /*压栈时，一次压入一个字符*/ if(ch==‘ ‘||ch==‘\n‘||ch==EOF) { while(!s.empty()) { printf("%c",s.top()); s.pop(); /*清除栈顶元素*/ } if(ch==‘\n‘||ch==EOF) break; /*绝对不能少，控制输出结束*/ printf(" "); } else s.push(ch); } printf("\n"); } return 0;}
hdu 1062 Text Reverse
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