C++类中的has-a关系和is-a关系

has-a关系是指类中包含了其他类的对象作为自己的成员变量，表示一个类拥有另外一个类的实例。比如一个汽车类可以包含一个引擎类的对象作为自己的成员变量，表示汽车拥有一个引擎。

is-a关系是指类之间的继承关系，表示一个类是另外一个类的一种特殊形式。比如一个圆形类可以继承自一个图形类，表示圆形是一种图形。在C++中，使用关键字“extends”来表示继承关系。

相关问题

sudo ./bootstrap --------------------------------------------- CMake 3.8.1, Copyright 2000-2017 Kitware, Inc. and Contributors Found GNU toolchain C compiler on this system is: gcc C++ compiler on this system is: g++ Makefile processor on this system is: make g++ is GNU compiler g++ has setenv g++ has unsetenv g++ does not have environ in stdlib.h g++ has stl wstring g++ has <ext/stdio_filebuf.h> --------------------------------------------- make: 'cmake' is up to date. CMake Error: The current CMakeCache.txt directory /home/GPU_Project/ARM-Mali-G72/Software_packet/DDK-Compile_Tool/cmake/cmake-3.8.1/CMakeCache.txt is different than the directory /home/test/GPU_Project/ARM-Mali-G72/Software_packet/DDK-Compile_Tool/cmake/cmake-3.8.1 where CMakeCache.txt was created. This may result in binaries being created in the wrong place. If you are not sure, reedit the CMakeCache.txt --------------------------------------------- Error when bootstrapping CMake: Problem while running initial CMake ---------------------------------------------

这个错误信息提示是在运行 CMake 过程中出现了问题，可能是由于 CMakeCache.txt 文件所在的目录不一致导致的。建议检查一下 CMakeCache.txt 文件所在的目录是否正确，并且尝试清空 CMake 缓存重新运行 bootstrap。你可以尝试以下命令：

rm -rf CMakeCache.txt
sudo ./bootstrap

如果还有问题，可以提供更多的错误信息，以便更好地帮助你解决问题。

Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is yes, if 6 is a decimal number and 110 is a binary number. Now for any pair of positive integers N1​ and N2​, your task is to find the radix of one number while that of the other is given. Input Specification: Each input file contains one test case. Each case occupies a line which contains 4 positive integers: N1 N2 tag radix Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set { 0-9, a-z } where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number radix is the radix of N1 if tag is 1, or of N2 if tag is 2. Output Specification: For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print Impossible. If the solution is not unique, output the smallest possible radix. 用c++写代码 并且要考虑radix大于36的情况

您想解决一个关于进制转换的问题，其中需要确定一个数字的进制，而另一个数字的进制已知。您希望用C++编写代码，并且要考虑进制大于36的情况。以下是一个可能的实现：

#include <iostream>
#include <string>
#include <cmath>

using namespace std;

int charToInt(char c) {
    if (isdigit(c)) {
        return c - '0';
    } else {
        return c - 'a' + 10;
    }
}

long long convertToDecimal(string num, int radix) {
    long long result = 0;
    int power = 0;
    for (int i = num.length() - 1; i >= 0; i--) {
        result += charToInt(num[i]) * pow(radix, power);
        power++;
    }
    return result;
}

int findRadix(string num, long long target) {
    char maxDigit = *max_element(num.begin(), num.end());
    int minRadix = charToInt(maxDigit) + 1; // Smallest possible radix
    
    // Binary search to find the radix
    int left = minRadix;
    int right = max(36, minRadix); // Consider radix larger than 36
    while (left <= right) {
        int mid = (left + right) / 2;
        long long converted = convertToDecimal(num, mid);
        if (converted == target) {
            return mid;
        } else if (converted < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    
    return -1; // Impossible
}

int main() {
    string N1, N2;
    int tag, radix;
    cin >> N1 >> N2 >> tag >> radix;
    
    if (tag == 1) {
        long long decimalN2 = convertToDecimal(N2, radix);
        int resultRadix = findRadix(N1, decimalN2);
        if (resultRadix == -1) {
            cout << "Impossible" << endl;
        } else {
            cout << resultRadix << endl;
        }
    } else {
        long long decimalN1 = convertToDecimal(N1, radix);
        int resultRadix = findRadix(N2, decimalN1);
        if (resultRadix == -1) {
            cout << "Impossible" << endl;
        } else {
            cout << resultRadix << endl;
        }
    }
    
    return 0;
}

这个程序首先定义了一个`charToInt`函数，用于将字符转换为对应的整数值。然后，`convertToDecimal`函数将给定进制的数字转换为十进制数。最后，`findRadix`函数使用二分查找来确定另一个数字的进制。在主函数中，根据输入的`tag`值选择相应的计算方法，并输出结果。

希望这可以帮助到您！如果您有任何其他问题，请随时问我。