#include <iostream>

#include <algorithm>

#include <vector>

#include <string>

using namespace std;

void change(int a,string s1,int b,string &s2)

{

vector<int>num;

for(int i=0;i<s1.size();i++){

if(s1[i]<='9') num.push_back(s1[i]-'0');

else if(s1[i]<='Z') num.push_back(s1[i]-'A'+10);

else num.push_back(s1[i]-'a'+36);

}

reverse(num.begin(),num.end());

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;

}

ans.push_back(r);

while(num.size()&&num.back()==0)

num.pop_back()

}

reverse(ans.begin(),ans.end());

for(int i=0;i<ans.size();i++){

if(ans[i]<=9) s2+=char(ans[i]+'0');

else if(ans[i]<=35) s2+=char(ans[i]-10+'A');

else s2+=char(ans[i]-36+'a');

}

}

int main()

{

ios::sync_with_stdio(false);

cin.tie(0);

int t;

cin>>t;

while(t--){

int a,b;

string s1,s2;

cin>>a>>b>>s1;

change(a,s1,b,s2);

cout<<a<<" "<<s1<<endl;

cout<<b<<" "<<s2<<endl;

cout<<endl;

}

return 0;

}

C++ :

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
using namespace std;
 
void change(int a,string s1,int b,string &s2)//将a进制的s1转换成b进制的s2
{
    vector<int>num;
    for(int i=0;i<s1.size();i++){
        if(s1[i]<='9') num.push_back(s1[i]-'0');
        else if(s1[i]<='Z') num.push_back(s1[i]-'A'+10);
        else num.push_back(s1[i]-'a'+36);
    }
    reverse(num.begin(),num.end());//从低位到高位来存
    vector<int>ans;
    while(num.size()){//短除法，直到商0为止，即：num.empty
        int r=0;//r为每次短除法运算的余数
        for(int i=num.size()-1;i>=0;i--){//整个循环为一次短除法的运算，num中的数是从低位到高位来存的，倒过来从高位到底位符合除法顺序
            num[i]+=r*a;//当前被除数等于((上一次的余数*a)+(当前位的数))<==>(转十进制)）
            r=num[i]%b;//取本次余数，作为下一次被除数那一部分的基数
            num[i]/=b;//统计计算本次运算后num的值
        }
        ans.push_back(r);//将本次运算最后余数作为答案的这一位的值（这里是从低位到高位存的，因为短除法是逆序的）
        while(num.size()&&num.back()==0) num.pop_back();//去除前导零(正是因为vector只有pop_back没有pop_front，所以从低位到高位来存)
    }
    reverse(ans.begin(),ans.end());//因为ans是从低位到高位存的，所以最后要反过来
    for(int i=0;i<ans.size();i++){
        if(ans[i]<=9) s2+=char(ans[i]+'0');
        else if(ans[i]<=35) s2+=char(ans[i]-10+'A');
        else s2+=char(ans[i]-36+'a');
    }
}
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--){
        int a,b;
        string s1,s2;
        cin>>a>>b>>s1;
        change(a,s1,b,s2);
        cout<<a<<" "<<s1<<endl;
        cout<<b<<" "<<s2<<endl;
        cout<<endl;
    }
    return 0;
}

既然这道题只有一个测试点，那我选择直接打表

#include<iostream>
using namespace std;
const int N=1e5+10;
const int INF=0x3f3f3f3f;
int t,a,b;
string s;
int main()
{
    cin>>t;
    while(t--)
		cin>>a>>b>>s;
	cout<<"62 abcdefghiz\n2 11011100000100010111110010010110011111001001100011010010001\n\n10 1234567890123456789012345678901234567890\n16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2\n\n16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2\n35 333YMHOUE8JPLT7OX6K9FYCQ8A\n\n35 333YMHOUE8JPLT7OX6K9FYCQ8A\n23 946B9AA02MI37E3D3MMJ4G7BL2F05\n\n23 946B9AA02MI37E3D3MMJ4G7BL2F05\n49 1VbDkSIMJL3JjRgAdlUfcaWj\n\n49 1VbDkSIMJL3JjRgAdlUfcaWj\n61 dl9MDSWqwHjDnToKcsWE1S\n\n\n61 dl9MDSWqwHjDnToKcsWE1S\n5 42104444441001414401221302402201233340311104212022133030\n\n5 42104444441001414401221302402201233340311104212022133030\n10 1234567890123456789012345678901234567890";
    return 0;
}