UVALive 3621 Power Calculus n次幂最少需要几次乘除得到 搜索

题目链接：点击打开链接

题意:

输入一个n(n<=1000)

问：有一个x，最少需要几次乘除可以算出x^n。

思路：

记忆化搜索

a[i] = j 表示i次幂需要j步得到

import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.TreeSet;
import java.util.Queue;

public class Main {
	static int N = 1000050;
	int[] a = new int[N];
	boolean dfs(int n, int num, int step){//用step步是否能得到n次,当前步数为num步
		if(num > step)return false;
		if(a[num] == n)return true;
		if((a[num] << (step-num)) < n)//一直平方也达不到n
			return false;
		for(int i = 0; i <= num; i++){
			a[num+1] = a[num]+a[i];
			if(a[num+1] <= 10000 && dfs(n,num+1,step))
				return true;
			a[num+1] = a[num] - a[i];
			if(a[num+1] > 0 && dfs(n,num+1, step))
				return true;
		}
		return false;
	}
	void work() {
		while(true){
			int n = cin.nextInt();
			if(n == 0)break;
			int i;
			for(i = 0; ; i++){
				a[0] = 1;
				if(dfs(n, 0, i))break;
			}
			out.println(i);
		}
	}
	Main() {
		cin = new Scanner(System.in);
		out = new PrintWriter(System.out);
	}

	public static void main(String[] args) {
		Main e = new Main();
		e.work();
		out.close();
	}

	public Scanner cin;
	public static PrintWriter out;
}