Word Reversal 【字符串处理】 ACM

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：将一段文章里的每一个单词逆置，但单词顺序不变。
解题分析： 【Word Reversal 【字符串处理】】一行一行的读取，读取完后将里边的每一个单词逆置后直接输出。
AC代码：

#include #include#define N 1000void reverse(char *x) { int i, tem, len = strlen(x); int n = len/2; for(i = 0; i < n; i++){ tem = x[i]; x[i] = x[len-1-i]; x[len-1-i] = tem; } return ; }int main() { char str[N], s[N]; int i, j, k, t, n, len, flag; scanf("%d", &t); while(t--){ scanf("%d", &n); getchar(); for(i = 0; i < n; i++){ gets(str); len = strlen(str); for(j = k = flag = 0; j <= len; j++){ if(str[j] == ' ' || str[j] == '\0'){ s[k] = '\0'; reverse(s); if(!flag){ printf("%s", s); flag = 1; } else printf(" %s", s); k = 0; continue; } s[k++] = str[j]; } printf("\n"); } if(t) printf("\n"); } return 0; }