ZOJ——1151 Word Reversal-蒲公英云

ZOJ——1151 Word Reversal

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1151

题目：

Word Reversal

Time 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

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;
}