万能题解（进制大多都能用！！！）

本人写了1个函数

//Base_Transfer(int m,string s,int k) //将m进制数s转化为k进制(返回值string)
//m <= 16 && k <= 16;
string Base_Transfer(int m,string s,int k)
{
	string c = "0123456789ABCDEF";
    long long val = 0;
    long long mul = 1;
    for(int i = s.size() - 1; i >= 1; i--)
    {
        if(s[i] >= '0' && s[i] <= '9')
        {
            val += int(s[i] - '0') * mul;
        }
        else
        {
            val += (int(s[i]) - 55) * mul;
        }
        mul *= m;
    }
    if(s[0] >= '0' && s[0] <= '9')
    {
        val += int(s[0] - '0') * mul;
    }
    else
    {
        val += (int(s[0]) - 55) * mul;
    }
    string res = "";
    while(val > 0)
    {
        res = c[val % k] + res;
        val = val / k;
    }
    return res;
}

原理很简单，把其他进制转10进制，再把10进制转其他进制。

所以答案很简单了……

#include <bits/stdc++.h>
using namespace std;

string BT(int m,string s,int k)
{
	string c = "0123456789ABCDEF";
    long long val = 0;
    long long mul = 1;
    for(int i = s.size() - 1; i >= 1; i--)
    {
        if(s[i] >= '0' && s[i] <= '9')
        {
            val += int(s[i] - '0') * mul;
        }
        else
        {
            val += (int(s[i]) - 55) * mul;
        }
        mul *= m;
    }
    if(s[0] >= '0' && s[0] <= '9')
    {
        val += int(s[0] - '0') * mul;
    }
    else
    {
        val += (int(s[0]) - 55) * mul;
    }
    string res = "";
    while(val > 0)
    {
        res = c[val % k] + res;
        val = val / k;
    }
    return res;
}
int main()
{
    string num;
    cin >> num;
    cout << BT(2,num,10) << endl << BT(8,num,10) << endl << BT(16,num,10);
}

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