原题链接

将一个 a 进制数转换成一个 b 进制数。思路在于模拟一个 a 进制数直接对一个十进制数 b 做除法。

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#include <iostream>
using namespace std;
#include <algorithm>

int n;

int main() {
  cin >> n;
  while (n--) {
    int a, b;
    string a_line, b_line;
    cin >> a >> b >> a_line;
    vector<int> num;
    for (auto &x : a_line) {
      if (x >= 'A' && x <= 'Z') {
        num.push_back(x - 'A' + 10);
      } else if (x >= 'a' && x <= 'z') {
        num.push_back(x - 'a' + 36);
      } else {
        num.push_back(x - '0');
      }
    }
    reverse(num.begin(), num.end());
    vector<int> res;
    while (num.size()) {
      int r = 0;
      for (int i = num.size() - 1; i >= 0; i--) {
        num[i] += r * a;
        r = num[i] % b;
        num[i] /= b;
      }
      res.push_back(r);
      while (num.size() && num.back() == 0) num.pop_back();
    }
    reverse(res.begin(), res.end());
    for (auto &x : res) {
      if (x <= 9) b_line += char(x + '0');
      if (x >= 10 && x <= 35) b_line += char(x - 10 + 'A');
      if (x >= 36) b_line += char(x - 36 + 'a');
    }
    cout << a << ' ' << a_line << endl;
    cout << b << ' ' << b_line << endl;
    cout << endl;
  }
  return 0;
}