题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1151
题目:
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
题目描述:
简单的字符串处理,给出一行单词,把这里面的单词翻转后在输出。单词之间可能不止有一个空格,别问我怎么知道的,,代码:
#include<stdio.h>
#include<string.h>
void qe(char *aw)
{
int i;
for(i = strlen(aw) - 1;i >=0;i--)
putchar(aw[i]);
}
int main()
{
int t,i,j,k;
int a,b,c;
while(scanf("%d",&a)!=EOF){
k = 0;
while(a--){
if(k)
putchar('\n');
scanf("%d",&b);
char *f = NULL;
char we[10010],ww[100];
gets(we);
for(i = 0 ;i < b;i ++){
gets(we);
for(t = 0,f = we;f[0] != '\0';t ++){
if(f[0] != ' '){
sscanf(f,"%s",ww);
qe(ww);
f = f + strlen(ww);
}
else{
printf("%c",f[0]);
f = f + 1;
}
}
putchar('\n');
}
k++;
}
}
return 0;
}