题解 | #进制转换2#

#include<iostream>
#include<stack>
#include<string>
using namespace std;
int main()
{
    int m, n;
    cin >> m >> n;
    long long int value = 0;
    string str;
    cin >> str;
    //转为十进制
    for (int i = 0; i < str.size(); ++i) {
        int current = (str[i] >= '0' &&
            str[i] <= '9') ? int(str[i] - '0') :
            int(str[i] - 'A' + 10);
        value = value * m + current;
    }
    //转为n进制
    stack<char> ans;
    while (value) {
        ans.push(value % n < 10 ? char((value % n) + '0') :
            char((value % n) - 10 + 'a'));
        value /= n;
    }
    while (!ans.empty()) {
        cout << ans.top();
        ans.pop();
    }
}