AcWing 3438. Number system conversion

Find the conversion of any two non negative integers with different bases （22  Base number  ∼∼ 1616  Base number ）, The integer given is in int Within the scope of .

The symbols of different bases are （0,1,…,9,a,b,…,f0,1,…,9,a,b,…,f） perhaps （0,1,…,9,A,B,…,F0,1,…,9,A,B,…,F）

Input format

There is only one line of input , Contains three integers  a,n,b.a  It means the following  n  yes  a  Hexadecimal integer ,b  Express the desire to  a  Hexadecimal integer  nn  convert to  bb  Hexadecimal integer .

a,ba,b  It's a decimal integer .

Data may contain leading zeros .

Output format

The output contains a line , This line has an integer for the converted  bb  Hexadecimal number .

When output, all alphabetic symbols are in upper case , namely （0,1,…,9,A,B,…,F0,1,…,9,A,B,…,F）.

Data range

2≤a,b≤162≤a,b≤16,
A given  a  Hexadecimal integer  n  The value range in decimal system is  [1,2147483647][1,2147483647].

sample input ：

15 Aab3 7

sample output ：

210306

  The main idea of the topic ：

take a Hexadecimal integer n Convert to b Hexadecimal integer

  Ideas ：

With 10 Base centered First the a Base to zero 10 Base number Then convert the converted decimal system to b Base number

  Any hexadecimal number is converted to 10 Base number

int toTen(int a, string num)   //a Hexadecimal number num Convert to 10 Base number 
{
    int ans = 0;
    int n = num.size();
    for (int i = 0; i < num.size(); i++) {
        if (isupper(num[i]))
            ans += pow(a, n - i - 1) * (num[i] - 'A' + 10);
        else if (islower(num[i]))
            ans += pow(a, n - i - 1) * (num[i] - 'a' + 10);
        else if (isdigit(num[i]))
            ans += pow(a, n - i - 1) * (num[i] - '0');
    }
    return ans;
}

  Decimal to arbitrary base

unordered_map<int, char> mp = {
        {0,  '0'}, {1,  '1'}, {2,  '2'}, {3,  '3'},
        {4,  '4'}, {5,  '5'}, {6,  '6'}, {7,  '7'},
        {8,  '8'}, {9,  '9'}, {10, 'A'}, {11, 'B'},
        {12, 'C'}, {13, 'D'}, {14, 'E'}, {15, 'F'}
};

string toB(int num, int b) {  //10 Hexadecimal number num Convert to b Base number 
    string ans;
    while(num >= b) {
        ans +=  mp[num % b];
        num /= b;
    }
    if(num != 0) ans += mp[num];
    reverse(ans.begin(), ans.end());
    return ans;
}

 AC Code 1：

#include <iostream>
#include <unordered_map>
#include <algorithm>
#include <cmath>

using namespace std;

unordered_map<int, char> mp = {
        {0,  '0'}, {1,  '1'}, {2,  '2'}, {3,  '3'},
        {4,  '4'}, {5,  '5'}, {6,  '6'}, {7,  '7'},
        {8,  '8'}, {9,  '9'}, {10, 'A'}, {11, 'B'},
        {12, 'C'}, {13, 'D'}, {14, 'E'}, {15, 'F'}
};

int toTen(int a, string num)   //a Hexadecimal number num Convert to 10 Base number 
{
    int ans = 0;
    int n = num.size();
    for (int i = 0; i < num.size(); i++) {
        if (isupper(num[i]))
            ans += pow(a, n - i - 1) * (num[i] - 'A' + 10);
        else if (islower(num[i]))
            ans += pow(a, n - i - 1) * (num[i] - 'a' + 10);
        else if (isdigit(num[i]))
            ans += pow(a, n - i - 1) * (num[i] - '0');
    }
    return ans;
}

string toB(int num, int b) {  //10 Hexadecimal number num Convert to b Base number 
    string ans;
    while(num) {
        ans +=  mp[num % b];
        num /= b;
    }
    reverse(ans.begin(), ans.end());
    return ans;
}

int main() {
    int a, b;
    string n;
    cin >> a >> n >> b;
    int num = toTen(a, n);
    cout << toB(num, b);
    return 0;
}

 

Concise Edition ：

/*
* @Author: Spare Lin
* @Project: AcWing2022
* @Date: 2022/6/30 21:26
* @Description: 3438.  Number conversion   source ： Peking University postgraduate entrance examination machine test questions 
*/

#include <iostream>
#include <algorithm>

using namespace std;

int main() {
    int a, b;
    string n;
    cin >> a >> n >> b;
    int res = 0;    //a Conversion from decimal to decimal 
    for (auto &x: n) {
        if(isupper(x)) res = res * a + x - 'A' + 10;
        if(islower(x)) res = res * a + x - 'a' + 10;
        if(isdigit(x)) res = res * a + x - '0';
    }
    string ans;     // Decimal to b Base number 
    while (res) {
        int tmp = res % b;
        if(tmp >= 10)
            ans += char(tmp + 'A' - 10);
        else
            ans += char(tmp + '0');
        res /= b;
    }
    reverse(ans.begin(), ans.end());
    cout << ans << '\n';
    return 0;
}