题解 | #进制转换2#

#include <iostream>
#include<string>
#include<vector>
using namespace std;
//将M进制的数转换为N进制的数输出
//M>=2 N<=36
//本题注意如果进制大于10时，用字符来表示
int chatToInt(char a) {
    if (a >= 'A' && a <= 'Z') {
        return a - 'A' + 10;
    } else {
        return a - '0';
    }
}
char intToChar(int x) {
    if (x >= 10) {
        return x - 10 + 'a';
    } else {
        return x + '0';
    }
}
int main() {
    int M, N;
    while (scanf("%d %d", &M, &N) != EOF) {
        string str;
        cin >> str; //不要endl
        long long x = 0;
        for (int i = 0; i < str.size(); i++) { //str.size()

            x = (chatToInt(str[i])) * 1 + x * M;
        }
        int j = 0;
        vector<char> answer;
        while ((int)(x / N) != 0) {

            char a = intToChar(x % N);
            answer.push_back(a);
            x /= N;
        }
        answer.push_back(intToChar(x % N));
        for (int i = answer.size() - 1; i >= 0; i--) {
            char b = answer[i]; //直接输出第几个，不要pop了
            cout << b;
        }
    }
}
// 64 位输出请用 printf("%lld")