题解 | #数制转换#

很丑但是很直观的解法

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

vector<int> res;

int main()
{
    string n;
    int a, b;
    cin >> a >> n >> b;
    long long x = 0;
    while (n.size() > 1 && n.front() == '0') //去除前导0
        n = n.substr(1);
    for (int i = 0; i < n.size(); ++i) //转成十进制了
        if (n[i] >= '0' && n[i] <= '9')
            x = x * a + n[i] - '0';
        else if (n[i] >= 'a' && n[i] <= 'f')
            x = x * a + n[i] - 'a' + 10;
        else if (n[i] >= 'A' && n[i] <= 'F')
            x = x * a + n[i] - 'A' + 10;
    while (x) //除留余数
    {
        res.push_back(x % b);
        x /= b;
    }
    for (int i = res.size() - 1; i >= 0; --i)
    {
        if (res[i] >= 10)
            printf("%c", 'A' + res[i] - 10);
        else
            cout << res[i];
    }
    
    return 0;
}