题解 | 数制转换

#include <iostream>
#include <string>
#include <cmath>
using namespace std;

int main() {
    int a, b;
    string str, res;
    long temp;
    int count;
    int t;
    while (cin >> a >> str >> b) {
        temp = 0;
        count = 0;
        for (int i = str.size() - 1; i >= 0; i--) {
            if (str[i] >= 'A' && str[i] <= 'Z') {
                t = str[i] - 'A' + 10;
            } else if (str[i] >= 'a' && str[i] <= 'z') {
                t = str[i] - 'a' + 10;
            } else {
                t = str[i] - '0';
            }
            temp += t * pow(a, count);
            count++;
        }
        count = 0;
        int i = 0;
        while (temp) {
            t = temp % b;
            temp /= b;
            if (t > 9) {
                switch (t) {
                    case 10:
                        res.push_back('A');
                        break;
                    case 11:
                        res.push_back('B');
                        break;
                    case 12:
                        res.push_back('C');
                        break;
                    case 13:
                        res.push_back('D');
                        break;
                    case 14:
                        res.push_back('E');
                        break;
                    case 15:
                        res.push_back('F');
                        break;
                }
            } else {
                res.push_back(t + '0');
            }
        }
        for (int j = res.size() - 1; j >= 0; j--) {
            cout << res[j];
        }
    }
}
// 64 位输出请用 printf("%lld")