[BOJ][2745] 진법 변환

1. 문제 링크

https://www.acmicpc.net/problem/2745

2. 사용 알고리즘

수학

3. 풀이

B진법 수 N을 10진수로 변환

EX) 3진법 102를 10진수로 변환하는 경우

• ans = 1

• ans = (1 * 3) + 0

• ans = (1 * 3 + 0) * 3 + 2

4. 소스 코드

4-1. C++

https://github.com/dev-aiden/problem-solving/blob/main/boj/2745.cpp

#include <iostream>

using namespace std;

int main(void) {
    ios_base::sync_with_stdio(false);
    string n; int b; cin >> n >> b;
    int len = n.length();
    int ans = 0;
    for (int i = 0; i < len; ++i) {
        int num = n[i] >= 65 ? n[i] - 55 : n[i] - 48;
        ans *= b;
        ans += num;
    }
    cout << ans << "\n";
    return 0;
}

4-2. JAVA

https://github.com/dev-aiden/problem-solving/blob/main/boj/2745.java

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int b = Integer.parseInt(st.nextToken());
        StringBuilder ans = new StringBuilder();
        while (n != 0) {
            int temp = n % b;
            if (temp < 10) ans.append((char)(temp + 48));
            else ans.append((char)(temp + 55));
            n /= b;
        }
        ans.reverse();
        System.out.println(ans);
    }
}