任意进制计算器 c++

以下是一个简单的 C++ 代码示例，用于将任意进制的数字转换为十进制，并实现基本的加法、减法、乘法和除法运算。

#include <iostream>
#include <string>

using namespace std;

// 将 n 进制的数 s 转换为十进制
int toDecimal(string s, int n) {
    int res = 0;
    for (int i = 0; i < s.length(); i++) {
        if (s[i] >= '0' && s[i] <= '9') {
            res = res * n + (s[i] - '0');
        } else {
            res = res * n + (s[i] - 'A' + 10);
        }
    }
    return res;
}

// 将十进制数 x 转换为 n 进制
string toBase(int x, int n) {
    string res = "";
    while (x) {
        int r = x % n;
        if (r <= 9) {
            res = char(r + '0') + res;
        } else {
            res = char(r - 10 + 'A') + res;
        }
        x /= n;
    }
    return res;
}

// 实现任意进制数的加法
string add(string a, string b, int n) {
    int carry = 0;
    int i = a.length() - 1, j = b.length() - 1;
    string res = "";
    while (i >= 0 || j >= 0 || carry) {
        int x = (i >= 0 ? a[i--] - '0' : 0);
        int y = (j >= 0 ? b[j--] - '0' : 0);
        int sum = x + y + carry;
        carry = sum / n;
        sum %= n;
        res = char(sum + '0') + res;
    }
    return res;
}

// 实现任意进制数的减法
string sub(string a, string b, int n) {
    int borrow = 0;
    int i = a.length() - 1, j = b.length() - 1;
    string res = "";
    while (i >= 0 || j >= 0) {
        int x = (i >= 0 ? a[i--] - '0' : 0);
        int y = (j >= 0 ? b[j--] - '0' : 0);
        int diff = x - y - borrow;
        if (diff < 0) {
            diff += n;
            borrow = 1;
        } else {
            borrow = 0;
        }
        res = char(diff + '0') + res;
    }
    // 去掉前导零
    while (res.length() > 1 && res[0] == '0') {
        res.erase(0, 1);
    }
    return res;
}

// 实现任意进制数的乘法
string mul(string a, string b, int n) {
    string res = "";
    int len = a.length() + b.length();
    int *c = new int[len];
    for (int i = 0; i < len; i++) {
        c[i] = 0;
    }
    for (int i = a.length() - 1; i >= 0; i--) {
        for (int j = b.length() - 1; j >= 0; j--) {
            int x = a[i] - '0';
            int y = b[j] - '0';
            int sum = x * y + c[i + j + 1];
            c[i + j + 1] = sum % n;
            c[i + j] += sum / n;
        }
    }
    int i = 0;
    while (i < len - 1 && c[i] == 0) {
        i++;
    }
    for (; i < len; i++) {
        res += char(c[i] + '0');
    }
    return res;
}

// 实现任意进制数的除法
string div(string a, string b, int n) {
    string res = "";
    int m = a.length(), n = b.length();
    int *A = new int[m];
    int *B = new int[n];
    int *C = new int[m];
    int *R = new int[n];
    for (int i = 0; i < m; i++) {
        A[i] = a[i] - '0';
        C[i] = 0;
    }
    for (int i = 0; i < n; i++) {
        B[i] = b[i] - '0';
    }
    for (int i = m - n; i >= 0; i--) {
        int d = 0;
        while (compare(A + i, B, n) >= 0) {
            sub(A + i, B, n);
            d++;
        }
        C[i] = d;
    }
    int i = 0;
    while (i < m && C[i] == 0) {
        i++;
    }
    for (; i < m; i++) {
        res += char(C[i] + '0');
    }
    return res;
}

// 比较两个任意进制数的大小
int compare(int *A, int *B, int n) {
    for (int i = 0; i < n; i++) {
        if (A[i] > B[i]) {
            return 1;
        } else if (A[i] < B[i]) {
            return -1;
        }
    }
    return 0;
}

int main() {
    string s = "1010";
    int n = 2;
    cout << toDecimal(s, n) << endl;

    int x = 10;
    int m = 2;
    cout << toBase(x, m) << endl;

    string a = "1101";
    string b = "101";
    cout << add(a, b, 2) << endl;

    cout << sub(a, b, 2) << endl;

    cout << mul(a, b, 2) << endl;

    cout << div(a, b, 2) << endl;

    return 0;
}

请注意，此代码示例仅适用于非负整数和较小的数字。在实际应用中，您可能需要考虑更多的边界情况和优化。