Question

我有以下经文：

A swarm of bees in May
Is worth a load hey;
A swarm of bees in June
Is worth a silver spoon;
A swarm of bees in July
Is hot a worth a fly.

我必须修改此文本，以便所有行以相同的位置结束。使用空格填充字符串中的位置数量不足。这些空间必须均匀分配我知道我的代码非常笨重，但我必须在代码中使用＆＃34;结构＆＃34;

如何找到最长的字符串并在其他字符串中添加空格来执行任务？

谢谢！

#include "stdafx.h"
#include "iostream"
#include <string.h> 
using namespace std;

struct VERSE {
    char row_one[25];
    char row_two[25];
    char row_three[25];
    char row_four[25];
    char row_five[25];
    char row_six[25];
};

int _tmain(int argc, _TCHAR* argv[])
{
    struct VERSE v;
    strcpy_s(v.row_one, "A swarm of bees in May");
    strcpy_s(v.row_two, "Is worth a load hey;");
    strcpy_s(v.row_three, "A swarm of bees in June");
    strcpy_s(v.row_four, "Is worth a silver spoon;");
    strcpy_s(v.row_five, "A swarm of bees in July");
    strcpy_s(v.row_six, "Is hot a worth a fly.");
    cout << v.row_one << endl << v.row_two << endl << v.row_three << endl
        << v.row_four << endl << v.row_five << endl << v.row_six << endl;

    cout << strlen(v.row_one) << endl;
    cout << strlen(v.row_two) << endl;
    cout << strlen(v.row_three) << endl;
    cout << strlen(v.row_four) << endl;
    cout << strlen(v.row_five) << endl;
    cout << strlen(v.row_six) << endl;

    //the length of row
    /*
    int length = 0;
    for(int i = 0; v.row_two[i] != '\0'; i++) {
        length++;
    }
    printf("Length of second row is: %d\n", length);
    */

    return 0;
}

Answer 1

A swarm of bees in May
Is worth a load hey;
A swarm of bees in June
Is worth a silver spoon;
A swarm of bees in July
Is hot a worth a fly.

让我烦恼。我想把它重写为：

A swarm of bees in May
Is worth a load of hay;
A swarm of bees in June
Is worth a silver spoon;
A swarm of bees in July
Is not worth a fly.

无论如何，在那之后：

我正在写这个答案，假设这是一个作业并且你被告知要使用c_strings。如果不是这种情况，那么使用std::string会更容易。

无论如何，我已经提出了这个代码：

#include "iostream"
#include <string.h> 
using namespace std;

const int maxrowlength = 25, maxrowcount = 6;

struct VERSE {
    char rows[maxrowcount][maxrowlength];
    int spaces[maxrowcount];
    int line_length[maxrowcount];
};

char* get_row(VERSE &v, int row)
{
    return &v.rows[row][0];
}

int main()
{
    struct VERSE v;
    strcpy(get_row(v,0), "A swarm of bees in May");
    strcpy(get_row(v,1), "Is worth a load hey;");
    strcpy(get_row(v,2), "A swarm of bees in June");
    strcpy(get_row(v,3), "Is worth a silver spoon;");
    strcpy(get_row(v,4), "A swarm of bees in July");
    strcpy(get_row(v,5), "Is hot a worth a fly.");

    //calculate lengths and count spaces
    int max_space_count = 0;
    for (size_t i = 0; i < maxrowcount; i++)
    {
        char* line = get_row(v,i);
        /*/we could find the length with strlen() and spaces with memchr() but 
           that will involve traversing the string multiple times (at least twice)
           we can do better
        /*/
        v.line_length[i] = 0;
        v.spaces[i] = 0;

        while(*line)
        {
            v.line_length[i]++;
            if(*line == ' '){v.spaces[i]++;}
            line++;
        }
        if (v.line_length[i] > max_space_count){max_space_count = v.line_length[i];}
    }

    for (size_t i = 0; i < maxrowcount; i++)
    {
        int length_diff = max_space_count - v.line_length[i];
        int spaces_to_add = v.spaces[i]?length_diff / v.spaces[i]:0; //number of spaces to add every word
        int extra_spaces  = v.spaces[i]?length_diff % v.spaces[i]:0; //extra spaces to add to make line fit
        char output[maxrowlength];
        char* current_output = output;
        char* current_word = get_row(v,i);
        char* current_word_end = current_word;

        while(*current_word)
        {
            current_word_end++;
            if (*current_word_end == ' ' || *current_word_end == '\0')
            {
                //write word to output
                strncpy(current_output, current_word, current_word_end - current_word);
                //update pointer to end of new word
                current_output += current_word_end - current_word;
                //write in the number of new spaces needed
                if (*current_word_end == ' ')
                {
                    for (int j = 0; j < spaces_to_add; j++)
                    {
                        *current_output = ' ';
                         current_output++;
                    }
                    //if extra spaces are needed, add those too
                    if (extra_spaces)
                    {
                        extra_spaces--;
                        *current_output = ' ';
                         current_output++;
                    }
                }
                //step current word to look at the next word
                current_word = current_word_end;
            }
        }
        //null terminate
        *current_output = '\0';
        strcpy(get_row(v,i),output);
    }

    for (size_t i = 0; i < maxrowcount; i++)
        {std::cout << get_row(v,i) << std::endl;}

    return 0;
}

输出：

A  swarm  of bees in May
Is  worth  a  load  hey;
A  swarm of bees in June
Is worth a silver spoon;
A  swarm of bees in July
Is  hot  a  worth a fly.

在此处查看： On Ideone

它的工作原理如下：

读入行
找出每一行的长度并计算其中的空格，找到最大长度
对于该行，计算所需的空格数，找出每个单词所需的数量，以及遗留的数量
将每个单词依次放入输出缓冲区
输入所需的空格数
如果行未完成则循环回4
null终止输出
循环回3
行已完成。

我是根据你的代码编写的，所以我可以从头开始做一些不同的事情：

line结构而不是您的VERSE
- 每个line包含其内容，空格数和长度
- line要存储在一个数组中 - 形成一节经文

这可能就是这个任务的目的，但是现在你的算法很有效（我不会为你做一切）;-)

Answer 2

您应该使用对象std::string来存储您的行。然后，使用函数std::max：

unsigned long int max_size = std::max(str1.length(),str2.length());
max_size = std::max(max_size, str3.length());
max_size = std::max(max_size, str4.length());
//...

使用＆＃34; struct＆＃34;修改文本。在C ++中

2 个答案: