BOSS战-普及综合练习3-数列_boss数列-优快云博客

本文解析了洛谷P1062题目，通过将数字转换为二进制形式并应用特定的数学运算来解决问题。使用Java实现算法，并提供完整代码示例。

题目链接：https://www.luogu.org/problemnew/show/P1062

题意理解

$1*3^0\\1*3^1+0*3^0\\1*3^1+1*3^0\\1*3^2+0*3^1+0*3^0$

然后就很显然，这是个二进制嘛，就把 $N$ 转换成二进制，然后乘上 $K$ 的对应次幂，然后求和。具体看代码。

代码

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

public class Main {
    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        int K = fs.nextInt();
        int N = fs.nextInt();
        long res = 0;
        String temp = Integer.toBinaryString(N);
        int len = temp.length();
        for(int i = 0; i < len; i++) {
            res += (temp.charAt(i) - '0') * Math.pow(K, len - i - 1);
        }
        System.out.println(res);
    }

    public static class FastScanner {
        private BufferedReader br;
        private StringTokenizer st;

        public void eat(String s) {
            st = new StringTokenizer(s);
        }

        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
            eat("");
        }

        public String nextLine() {
            try {
                return br.readLine();
            } catch (Exception e) {
                // TODO: handle exception
            } 
            return null;
        }

        public boolean hasNext() {
            while(!st.hasMoreTokens()) {
                String s = nextLine();
                if(s == null) {
                    return false;
                }
                eat(s);
            }
            return true;
        }

        public String nextToken() {
            hasNext();
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.valueOf(nextToken());
        }
    }
}