C++ 我如何简化这个程序（字谜）？

这是计算字符移位数的算法。我如何简化这一点

#include <cstdio>
#include <vector>
#include <iostream>
using namespace std;
#define SIZE 20
int w[1 << (SIZE + 1)];

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可以通过使用C++提供的算法来简化函数。 在下面的示例代码中，我们读取2个字符串，然后旋转其中一个，然后查看它是否等于另一个。我们将重复该操作，直到找到匹配项，或者直到检测到不可能进行转换。代码已注释，应该可以理解。如果没有，请询问

#include <iostream>
#include <string>
#include <algorithm>

int main()
{
    std::string string1{};  std::string string2{};
    std::cout << "Enter 2 strings with same number of digits:\n";
    std::cin >> string1 >> string2; // Read strings
    // Strings must have same size
    if (string1.size() == string2.size()) {
        // Countes the number of rotations to the right
        size_t rotateCounter{0};
        do  {
            // If rotated string is equal to original, then we found something
            if (string1 == string2) break;
            // Rotate right
            std::rotate(string1.rbegin(),string1.rbegin()+1,string1.rend());
            // We have done one more rotation
            ++rotateCounter;
        } while(rotateCounter < string1.size());
        // CHeck, if we could find a solution
        if ((rotateCounter == string1.size()) && (string1 != string2))  {
            std::cout << "Total different strings. No rotation transformation possible\n";
        } else {
            std::cout << "Number of right shifts needed: " << rotateCounter << '\n';
        }
    } else {
        std::cerr << "Size of strings not equal\n";
    }
    return 0;
}

最后但并非最不重要。该版本使用纯静态数组和手工制作的旋转算法

这应该足够了。也许下一次，非常明确的要求会有所帮助。然后我们可以选择正确的解决方案

#include <iostream>
#include <algorithm>
#include <iterator>

constexpr size_t MaxDigits = 1000000;

char LetterArray1[MaxDigits] {};
char LetterArray2[MaxDigits] {};

// Rotate array one char to the right
inline void rotateRight(char(&arrayToRotate)[MaxDigits], size_t numberOfLetters)
{
    --numberOfLetters;  
    char temp = arrayToRotate[numberOfLetters];
    for (size_t i = numberOfLetters; i > 0; --i) arrayToRotate[i] = arrayToRotate[i - 1];
    arrayToRotate[0] = temp;
}

int main() {
    // Get the number of letters that the user wants to use
    std::cout << "Enter the number of letters:   ";
    size_t numberOfLetters{ 0 }; std::cin >> numberOfLetters;
    // Check input for underflow or overflow
    if (numberOfLetters <= MaxDigits)
    {
        // Now read to strings from the console
        std::cout << "\nEnter  2 strings:\n";
        std::string inputString1{}; std::string inputString2{};
        // Read 2 strings
        std::cin >> inputString1 >> inputString2;
        // If user enters too short string, we would run into trouble. Therefore, pad string with spaces
        inputString1 += std::string(numberOfLetters, ' ');
        inputString2 += std::string(numberOfLetters, ' ');

        // Copy strings to array
        inputString1.copy(LetterArray1, numberOfLetters);
        inputString2.copy(LetterArray2, numberOfLetters);

        // So, now we have the 2 strings in our arrays
        // We will rotate Array1 and compare the result with Array 2. And we count the number of shifts
        bool matchFound{ false };
        size_t rotateCounter{ 0 };
        while (!matchFound && (rotateCounter < numberOfLetters))
        {
            if (0 == std::memcmp(LetterArray1, LetterArray2, numberOfLetters)) {
                matchFound = true; break;
            }
            rotateRight(LetterArray1, numberOfLetters);
            ++rotateCounter;
        }
        if (matchFound)     {
            std::cout << "\nNecessary rotations: " << rotateCounter << '\n';
        }
        else {
            std::cout << "\nNo Match Found\n";
        }
    }
    else {
        std::cerr << "***** Number of letters entered is to big. Max allowed: " << MaxDigits << '\n';
    }
    return 0;
}

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若你们有一个可以运行的程序，只是想知道如何改进它，那个么最好的地方就是简化可读性？使用有意义的变量名称在代码中添加一些说明算法工作原理的文档。欢迎使用StackOverflow。请按照创建此帐户时的建议，遵循帮助文档中的过帐指导原则，和在这里申请。StackOverflow不是设计、编码、研究或教程资源。请检查一般问题。好的，但我应该使用数组和变量N更改字母数。n<1000 000FWIW，需求中给出的最大值表明OP的老师希望不使用动态分配，而是使用问题中所示的静态存储阵列。我不会那么做的。但是嘿，噢，我插入n=4，inputString1=AACB和inputString2=BACA。结果未找到，但应输出：4.Hm？我不明白。如何通过旋转将AACB转换为BACA？逻辑是什么？请ss:AACB->BAAC->CBAA->ACBA->AACB

#include <iostream>
#include <algorithm>
#include <iterator>

constexpr size_t MaxDigits = 1000000;

char LetterArray1[MaxDigits] {};
char LetterArray2[MaxDigits] {};

// Rotate array one char to the right
inline void rotateRight(char(&arrayToRotate)[MaxDigits], size_t numberOfLetters)
{
    --numberOfLetters;  
    char temp = arrayToRotate[numberOfLetters];
    for (size_t i = numberOfLetters; i > 0; --i) arrayToRotate[i] = arrayToRotate[i - 1];
    arrayToRotate[0] = temp;
}

int main() {
    // Get the number of letters that the user wants to use
    std::cout << "Enter the number of letters:   ";
    size_t numberOfLetters{ 0 }; std::cin >> numberOfLetters;
    // Check input for underflow or overflow
    if (numberOfLetters <= MaxDigits)
    {
        // Now read to strings from the console
        std::cout << "\nEnter  2 strings:\n";
        std::string inputString1{}; std::string inputString2{};
        // Read 2 strings
        std::cin >> inputString1 >> inputString2;
        // If user enters too short string, we would run into trouble. Therefore, pad string with spaces
        inputString1 += std::string(numberOfLetters, ' ');
        inputString2 += std::string(numberOfLetters, ' ');

        // Copy strings to array
        inputString1.copy(LetterArray1, numberOfLetters);
        inputString2.copy(LetterArray2, numberOfLetters);

        // So, now we have the 2 strings in our arrays
        // We will rotate Array1 and compare the result with Array 2. And we count the number of shifts
        bool matchFound{ false };
        size_t rotateCounter{ 0 };
        while (!matchFound && (rotateCounter < numberOfLetters))
        {
            if (0 == std::memcmp(LetterArray1, LetterArray2, numberOfLetters)) {
                matchFound = true; break;
            }
            rotateRight(LetterArray1, numberOfLetters);
            ++rotateCounter;
        }
        if (matchFound)     {
            std::cout << "\nNecessary rotations: " << rotateCounter << '\n';
        }
        else {
            std::cout << "\nNo Match Found\n";
        }
    }
    else {
        std::cerr << "***** Number of letters entered is to big. Max allowed: " << MaxDigits << '\n';
    }
    return 0;
}