UVA10254-The Priest Mathematician)

UVA10254-The Priest Mathematician)

UVA10254-The Priest Mathematician)

Question Link

The tower of the four pillars.

Solution: There is a prompt in the question: First remove k with four pillars, and then remove the remaining n-k with the help of three pillars, then, move n to the column where n-k is located. Then F [n] = min (2 * F [k] + H [n-k]); H [n-k] = 2 ^ (n-k)-1; print out the previous 60 items and then print out F [n]-f [n-1]. The rule is found:
F [1] = 1;

F [2] = F [1] + 2 ^ 1;
F [3] = F [2] + 2 ^ 1; (2)

F [4] = f [3] + 2 ^ 2;
F [5] = f [4] + 2 ^ 2;
F [6] = f [5] + 2 ^ 2; (3)

F [7] = f [6] + 2 ^ 3;
... (4)

However, if n reaches 10000, a large number is required.

Code:

import java.util.*;import java.math.*;import java.io.*;public class Main {    static BigInteger f[] = new BigInteger[10005];    public static void init () {        f[0] = BigInteger.ZERO;        f[1] = BigInteger.valueOf(1);        int k = 1, j = 2;        BigInteger addnum;        while (j <= 10000) {            addnum = BigInteger.valueOf(1).shiftLeft(k);            for (int i = 0; i < k + 1 && j <= 10000; i++, j++)                 f[j] = f[j - 1].add(addnum);            k++;        }        }    public static void main(String args[]) {        Scanner cin = new Scanner(System.in);        init();        while (cin.hasNext()) {            int n = cin.nextInt();            System.out.println(f[n]);        }    }}