poj 3134 Power Calculus（迭代加深dfs+强剪枝）

Description

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

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

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

x2 = x × x, x3 = x2 × x, x6 = x3 × x3, x7 = x6 × x, x14 = x7 × x7, x15 = x14 × x, x30 = x15 × x15, x31 = x30 × 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, x4 = x2 × x2, x8 = x4 × x4, x8 = x4 × x4, x10 = x8 × x2, x20 = x10 × x10, x30 = x20 × x10, x31 = x30 × x.

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):

x2 = x × x, x4 = x2 × x2, x8 = x4 × x4, x16 = x8 × x8, x32 = x16 × 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 should be x to a positive integer’s power. In others words, x−, 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 . 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

Sample Output

Source

Japan 2006

求只用乘法和除法最快多少步可以求到x^n

其实答案最大13，但由于树的分支极为庞大在IDDFS的同时，我们还要加2个剪枝

1 如果当前序列最大值m*2^(dep-k)<n则减去这个分支

2 如果出现两个大于n的数则要减去分支。因为里面只有一个有用，我们一定可以通过另外更加短的路径得到答案

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 int num;
 int way[];
 bool dfs(int n,int step){
     if(num>step) return false;
     if(way[num]==n) return true;
     if(way[num]<<(step-num)<n) return false;//强剪枝
     for(int i=;i<=num;i++){
         num++;
         way[num]=way[num-]+way[i];
         if(way[num]<= && dfs(n,step)) return true;

         way[num]=way[num-]-way[i];
         if(way[num]> && dfs(n,step)) return true;
         num--;
     }
     return false;
 }
 int main()
 {
     int n;
     while(scanf("%d",&n)==){
         if(n==){
             break;
         }

         //迭代加深dfs
         int i;
         for(i=;;i++){
             way[num=]=;
             if(dfs(n,i))
                 break;
         }
         printf("%d\n",i);

     }
     return ;
 }