ZOJ 1151 Word Reversal
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：把输入的字符串以空格为界限进行逆置
解题思路：模拟一下整个过程即可
AC代码：
#include<stdio.h>
#include<string.h>
int main(void)
{
char a[1005];
int N, i, j, len, k, n;
scanf("%d", &N);
while(N--)
{
scanf("%d", &n);
getchar();
while(n --)
{
k = 0;
gets(a);
len = strlen(a);
for(i = 0; a[k] != '\0'; i ++)
{
if(i == 0)
{
j = i;
while(a[j] != ' ' && a[j] != '\0')
{
j ++;
}
if(a[j] == '\0')
{
k = j;
}
j = j - 1;
while(i <= j)
{
printf("%c", a[j]);
j --;
}
}
else
{
if(a[i] == ' ')
{
j = i;
j ++;
while(a[j] != ' ' && a[j] != '\0')
{
j ++;
}
while(i < j && a[j] != '\0')
{
printf("%c", a[j]);
j --;
}
if(a[j] == '\0')
{
k = j;
}
if(a[j] == '\0')
{
j --;
printf(" %c", a[j]);
while(i + 1 < j && a[j] != '\0')
{
j --;
printf("%c", a[j]);
}
}
}
}
}
printf("\n");
}
if(N != 0)
{
printf("\n");
}
}
return 0;
}
