IDA*_DFS

链接：3134 -- Power Calculus

Description

Starting with x and repeatedly multiplying by x, we can compute x31 with thirty multiplications:

The operation of squaring can be appreciably shorten the sequence of multiplications. The following is a way to compute x31 with eight multiplications:

This is not the shortest sequence of multiplications to compute x31. There are many ways with only seven multiplications. The following is one of them:

If division is also available, we can find a even shorter sequence of operations. It is possible to compute x31 with six operations (five multiplications and one division):

This is one of the most efficient ways to compute x31 if a division is as fast as a multiplication.

Your mission is to write a program to find the least number of operations to compute xn by multiplication and division starting with x for the given positive integer n. Products and quotients appearing in the sequence should be x to a positive integer’s power. In others words, x−3, for example, should never appear.

Input

The input is a sequence of one or more lines each containing a single integer n. n is positive and less than or equal to 1000. The end of the input is indicated by a zero.

Output

Your program should print the least total number of multiplications and divisions required to compute xn starting with x for the integer n. The numbers should be written each in a separate line without any superfluous characters such as leading or trailing spaces.

Sample Input

1
31
70
91
473
512
811
953
0

Sample Output

0
6
8
9
11
9
13
12

#include <iostream>
using namespace std;
int n;
int a[1024] = {0};
int dfs(int cur, int depth)
{
    if (cur == depth)
    {
        if (a[cur] == n)
            return 1;
        else
            return 0;
    }
    if (a[cur] << (depth - cur) < n) // 剪枝，如果剩下的步骤都是倍增还达不到
        return 0;
    for (int i = 0; i <= cur; i++) // 通过先前的值构造下一步的值
    {
        a[cur + 1] = a[cur] + a[i];
        if (dfs(cur + 1, depth))
            return 1;
        a[cur + 1] = a[cur] - a[i];
        if (dfs(cur + 1, depth))
            return 1;
    }
    return 0;
}
int main()
{
    // freopen("C:/Users/zhaochen/Desktop/input.txt", "r", stdin);
    cin.tie(0);
    cout.tie(0);
    a[0] = 1; // 从1开始
    while (cin >> n && n)
    {
        for (int depth = 0;; depth++) // 尝试计算用depth步从1得到n
        {
            if (dfs(0, depth))
            {
                cout << depth << endl;
                break;
            }
        }
    }
    return 0;
}