Word Reversal 【字符串处理】

素颜马尾好姑娘i 2022-05-31 01:26 93阅读 0赞

## 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： ####

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<stdio.h>
    #include<string.h>
    
    #define N 1000
    
    void 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;
    }