C++ 使用“修改文本”；结构"；在C++；

我有一首诗：

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.

我必须修改这个文本，使所有的行都以相同的位置结束。

---

使用空格的字符串补码中的位置数不足。这些空间必须平均分配 我知道我的代码非常庞大，但我必须在代码中使用“struct”

如何找到最长的字符串并在其他字符串中添加空格以执行任务

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;
}

#包括“stdafx.h”
#包括“iostream”
#包括
使用名称空间std；
结构诗{
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[]
{
结构诗v；
strcpy_s（v.row_one，“五月的蜂群”）；
strcpy_s（v.row_two，“值得一负荷嘿；”）；
strcpy_s（v.row_三，“六月的一群蜜蜂”）；
strcpy_s（v.row_四，“值一银匙；”）；
strcpy_s（v.row_five，“七月的一群蜜蜂”）；
strcpy_s（v.row_六，“热得值得一飞。”）；
cout您应该使用对象
std:：string
来存储行。
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());
//...

只是让我很恼火。我想把它改写为：

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.

不管怎么说，这件事已经解决了：

我在写这个答案时假设这是一个作业，并且你被告知要使用c_字符串。如果不是这样，那么使用
std:：string
会更容易

无论如何，我已经想出了以下代码：

#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;
}

请看这里：

它是这样工作的：

排队阅读
找出每行的长度并计算其中的空格，找出最大长度
对于行，计算所需的空格数，找出每个单词所需的空格数，以及剩余的空格数
将每个字依次放入输出缓冲区
放入所需数量的空格
如果行未完成，则返回到4
null终止输出
如果未处理完整输入，则循环回3
台词都写好了
我是根据您的代码编写的，因此有几件事我会从头开始做：


line
struct而不是
VERSE

每行
都包含其内容、空格数和长度

行
s存储在数组中-构成韵文


这可能就是这个任务的目的，但是现在你的算法运行起来应该很容易做到（我不会为你做所有的事情）不要使用交叉标记C和C++，语言也有很大的不同。你也应该使用<代码> STD::String < /Cord>代替C字符串。你可能还想使用<代码> STD:：vector < /C>而不是固定的“行”。然后，代码变得非常简单。“使用空格的字符串中的位置数不足”我不理解这个。也从代码< >代码>结构> <代码>，C++不需要。你不允许使用<代码> STD:：矢量< /代码>和<代码> STD:：String < /C> >，如果这是一个练习，你就别无选择，只能手动检查<代码>结构> <代码>的每个成员，以获得它的长度，得到最大值，然后做同样，这一次，在“行”中添加空格的数量更少。@NeilKirk：他希望在单词之间插入新的空格字符，以补充现有字符，从而使每个句子的最终长度相同。
#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;
}

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.