ZOJ-1151 Word Reversal

Word ReversalTime 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
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

题意概括：输入n行语句，由英文单词和空格构成，将所有的单词反转输出，空格不动。如To反转为oT输出。

解题思路：对输入的语句从头到尾一个循环，如果该位置的字符不是空格存入一个字符串中，遇到空格或回车符时倒着输出字符串中的单词。也可以用栈，非空格入栈，遇到空格栈里内容依次出栈。

代码：

栈#include<stdio.h>
#include<stack>
using namespace std;
#include<string.h>
int main()
{
int i,j,k,t,n;
char str[5010];
stack<char> s;
scanf("%d",&t);

while(t--)
{
scanf("%d",&n);
getchar();
for(i=0;i<n;i++)
{
memset(str,0,sizeof(str));
gets(str);
k=0;

for(j=0;;j++)
{
if(str[j]==' '||str[j]=='\0')
{
while(!s.empty())
{
char s1=s.top();
printf("%c",s1);
s.pop();
}
if(str[j]==' ')
printf(" ");

}
if(str[j]!=' ')
{
s.push(str[j]);
}

if(str[j]=='\0')
break;

}
s.pop();
printf("\n");
}
if(t!=0)
printf("\n");
}
return 0;
}
字符串#include<stdio.h>
#include<string.h>
int main()
{
int i
4000
,j,k,t,n;
char str[1010][1101],str2[200];
scanf("%d",&t);

while(t--)
{
scanf("%d",&n);
getchar();
for(i=0;i<n;i++)
{
memset(str,0,sizeof(str));
gets(str[i]);
k=0;
//printf("\n");
for(j=0;;j++)
{
if(str[i][j]==' '||str[i][j]=='\0')遇到空格或者回车符时把str2中存的单词倒着输出
{
int l=strlen(str2);
for(int u=l-1;u>=0;u--)
{
printf("%c",str2[u]);
}
k=0;
if(str[i][j]==' ')
printf(" ");
memset(str2,0,sizeof(str2));
}
if(str[i][j]!=' ')//将单词存入字符串str2中
{
str2[k]=str[i][j];
k++;
}

if(str[i][j]=='\0')
break;

}
memset(str2,0,sizeof(str2));
printf("\n");
}
if(t!=0)
printf("\n");
}
return 0;
}