Write a program, can realize a number from one base to another base.

There are 62 different digits {0-9,A-Z,a-z}.

Input format

Enter an integer for the first row to represent the number of rows to follow.

Next, each line contains three numbers, first input base (decimal), then output base (decimal), and finally input number represented by input base. The numbers are separated by spaces.

Both input and output are in the range of 2 to 62.

(in decimal) A = 10, B = 11 ，Z = 35，a = 36，b = 37，… , z = 61 (0-9 still means 0-9).

Output format

For each set of base conversion, the output of the program consists of three lines.

The first line contains two numbers, first the input base (decimal) and then the input number in the input base.

The second line contains two numbers, first the output base (decimal) and then the input number in the output base.

The third line is blank.

The numbers in the same line are separated by spaces.

Input example:

8
62 2 abcdefghiz
10 16 1234567890123456789012345678901234567890
16 35 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2
35 23 333YMHOUE8JPLT7OX6K9FYCQ8A
23 49 946B9AA02MI37E3D3MMJ4G7BL2F05
49 61 1VbDkSIMJL3JjRgAdlUfcaWj
61 5 dl9MDSWqwHjDnToKcsWE1S
5 10 42104444441001414401221302402201233340311104212022133030

Output example:

62 abcdefghiz
2 11011100000100010111110010010110011111001001100011010010001

10 1234567890123456789012345678901234567890
16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2

16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2
35 333YMHOUE8JPLT7OX6K9FYCQ8A

35 333YMHOUE8JPLT7OX6K9FYCQ8A
23 946B9AA02MI37E3D3MMJ4G7BL2F05

23 946B9AA02MI37E3D3MMJ4G7BL2F05
49 1VbDkSIMJL3JjRgAdlUfcaWj

49 1VbDkSIMJL3JjRgAdlUfcaWj
61 dl9MDSWqwHjDnToKcsWE1S

61 dl9MDSWqwHjDnToKcsWE1S
5 42104444441001414401221302402201233340311104212022133030

5 42104444441001414401221302402201233340311104212022133030
10 1234567890123456789012345678901234567890

Base conversion function:

void change(int a,string s1,int b,string &s2)//Convert s1 of a-Base to s2 of b-Base
{
    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());//From low to high
    vector<int>ans;
    while(num.size()){//Short division until quotient 0, i.e. num.empty
        int r=0;//r is the remainder of each short division operation
        for(int i=num.size()-1;i>=0;i--){//The whole cycle is a short division operation. The numbers in num are stored from the low to the high, and vice versa
            num[i]+=r*a;//Current divisor is equal to ((last remainder * a) + (number of current bits)) < = = > (decimal conversion))
            r=num[i]%b;//Take this remainder as the base of the next dividend
            num[i]/=b;//Count and calculate the value of num after this operation
        }
        ans.push_back(r);//Take the last remainder of this operation as the value of this bit of the answer (here it is stored from low to high, because the short division is in reverse order)
        while(num.size()&&num.back()==0) num.pop_back();//Remove leading zeros (just because vector has only pop back and no pop front, it stores from low to high)
    }
    reverse(ans.begin(),ans.end());//Because ans is stored from low to high, so in the end, reverse
    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');
    }
}

Full code:

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

void change(int a,string s1,int b,string &s2)//Convert s1 of a-Base to s2 of b-Base
{
    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());//From low to high
    vector<int>ans;
    while(num.size()){//Short division until quotient 0, i.e. num.empty
        int r=0;//r is the remainder of each short division operation
        for(int i=num.size()-1;i>=0;i--){//The whole cycle is a short division operation. The numbers in num are stored from the low to the high, and vice versa
            num[i]+=r*a;//Current divisor is equal to ((last remainder * a) + (number of current bits)) < = = > (decimal conversion))
            r=num[i]%b;//Take this remainder as the base of the next dividend
            num[i]/=b;//Count and calculate the value of num after this operation
        }
        ans.push_back(r);//Take the last remainder of this operation as the value of this bit of the answer (here it is stored from low to high, because the short division is in reverse order)
        while(num.size()&&num.back()==0) num.pop_back();//Remove leading zeros (just because vector has only pop back and no pop front, it stores from low to high)
    }
    reverse(ans.begin(),ans.end());//Because ans is stored from low to high, so in the end, reverse
    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;
}