UVA 1374 迭代加深搜索

http://vjudge.net/contest/view.action?cid=50643#problem/C

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

x² = x x x,      x³ = x² x x,      x⁴ = x³ x x,      ...  ,      x³¹ = x³⁰ x x.

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

x² = x x x,      x³ = x² x x,      x⁶ = x³ x x³,      x⁷ = x⁶ x x,      x¹⁴ = x⁷ x x⁷, 
x¹⁵ = x¹⁴ x x,      x³⁰ = x¹⁵ x x¹⁵,      x³¹ = x³⁰ x x.

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:

x² = x x x,      x⁴ = x² x x²,      x⁸ = x⁴ x x⁴,      x¹⁰ = x⁸ x x², 
x²⁰ = x¹⁰ x x¹⁰,      x³⁰ = x²⁰ x x¹⁰,      x³¹ = x³⁰ x x.

There however is no way to compute  x³¹ with fewer multiplications. Thus this is one of the most efficient ways to compute  x³¹ only by multiplications.

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

x² = x x x,      x⁴ = x² x x²,      x⁸ = x⁴ x x⁴,      x¹⁶ = x⁸ x x⁸,      x³² = x¹⁶ x x¹⁶,      x³¹ = x³² ÷ x.

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 of operations should be x to a positive integer's power. In other 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

题目大意：给出一个数n，求利用x如何在最短的步数内变成x^n,可以利用乘法和除法必须利用已经求出的x的幂指数，而且相除之后不能出现x的负次幂。

解题思路：最直接的想法就是利用dfs进行搜索，但是有一点我们在深搜的时候不知道什么时候是个结尾，因此我们可以利用限定搜索的层数。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
int limt_dept;
int n,num,vis[1020];
bool dfs()
{
    if(num>limt_dept) return false;
    if(vis[num]==n) return true;
    if(vis[num]<<(limt_dept-num)<n)return false;
    for(int i=0;i<=num;i++)
    {
        num++;
        vis[num]=vis[num-1]+vis[i];
        if(vis[num]<=2000&&dfs())return true;
        vis[num]=vis[num-1]-vis[i];
        if(vis[num]>0&&dfs())return true;
        num--;
    }
    return false;
}
int main()
{
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        limt_dept=0;
        while(limt_dept<=20)
        {
            num=0;
            vis[num]=1;
            if(dfs())
                break;
            limt_dept++;
        }
        printf("%d\n",limt_dept);
    }
    return 0;
}