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

3
olleh !dlrow
m'I morf .udh
I ekil .mca

 

Sample Output

hello world!
I'm from hdu.
I like acm.

Hint

Remember to use getchar() to read ' ' 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：

#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
const int maxn=1000+10;
char str[maxn];
int main()
{
    stack<char>s;
    int T;
    char ch;
    scanf("%d",&T);
    getchar();
    while(T--)
    {
        //s.clear;
        gets(str);
        //getchar();
        for(int i=0;i<strlen(str);i++)
        {
            if(str[i]!=' ')
            {
                s.push(str[i]);
            }
            if(str[i]==' '||i==(strlen(str)-1))
            {
                while(!s.empty())
                {
                    ch=s.top();
                    printf("%c",ch);
                    s.pop();
                }
                if(i!=(strlen(str)-1))
                printf(" ");
            }
        }
        printf("
");
    }
}

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=='
'||ch==EOF)
            {
                while(!s.empty())
                {
                    printf("%c",s.top()); 
                    s.pop(); /*清除栈顶元素*/
                }
                if(ch=='
'||ch==EOF)  
                    break;  /*绝对不能少，控制输出结束*/
                printf(" ");
            }
            else
                s.push(ch);
        }
        printf("
");
    }
    return 0;
}