POJ 3134 Power Calculus (iterative deepening dfs+ strong pruning)

Description

starting with X and repeatedly multiplying to X, we can compute x31 with thirty multiplications:x2= xxx, x3 = x2xx, x4 = x3xx, ..., x31 =x30xx.the operation of squaring can be appreciably shorten the sequence of multiplications. The following isA-compute x31 with eight multiplications:x2= xxx, x3 = x2xx, x6 = x3xx3, X7 = x6xx, x14 = x7xx7, x15 = x14xx, x30 = x15xx15, x31 =X30xx.this isNot the shortest sequence of multiplications to compute x31. There is many ways with only seven multiplications. The following isOne of them:x2= xxx, x4 = x2xx2, x8 = x4xx4, x8 = x4xx4, x10 = x8xx2, x20 = x10xx10, x30 = x20xx10, x31 =X30xx.if Division isAlso available, we can find a even shorter sequence of operations. It ispossible to compute X31 with six operations (five multiplications and one division): X2= xxx, x4 = x2xx2, x8 = x4xx4, x16 = x8xx8, x32 = x16xx16, x31 =X32÷x.this isOne of the most efficient ways to compute x31ifA division is  asFast asa multiplication. Your Mission isTo write a program to find the least number of operations to compute XN by multiplication and division starting with X forThe given positive integer n. Products and quotients appearinginchThe sequence should is x to a positive integer ' s power. In others words, x−3, forExample, should never appear.

Input

 is  is  +  is indicated by a zero.

Output

 for inch  as leading or trailing spaces.

Sample Input

1  to  -  the 473  + 811 953 0

Sample Output

0 6 8 9  One 9  -  A

Source

Japan 2006

Find out how many steps you can take with multiplication and division to get X^n

In fact, the answer is up to 13, but since the branches of the tree are very large in iddfs, we have to add 2 pruning

1 Subtract this branch if the current sequence maximum m*2^ (dep-k) <n

2 If there are two numbers greater than N, subtract the branch. Because there is only one useful in it, we can always get the answer by a shorter path.

1#include <iostream>2#include <cstdio>3#include <cstring>4 using namespacestd;5 intnum;6 intway[1006];7 BOOLDfsintNintStep) {8     if(Num>step)return false;9     if(way[num]==n)return true;Ten     if(way[num]<< (step-num) <n)return false;//Strong Pruning One      for(intI=0; i<=num;i++){ Anum++; -way[num]=way[num-1]+Way[i]; -         if(way[num]<= +&& DFS (N,step))return true; the  -way[num]=way[num-1]-Way[i]; -         if(way[num]>0&& DFS (N,step))return true; -num--; +     } -     return false; + } A intMain () at { -     intN; -      while(SCANF ("%d", &n) = =1){ -         if(n==0){ -              Break; -         } in  -         //Iterate deepen DFS to         inti; +          for(i=0;; i++){ -way[num=0]=1; the             if(Dfs (n,i)) *                  Break; $         }Panax Notoginsengprintf"%d\n", i); -          the     } +     return 0; A}

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POJ 3134 Power Calculus (iterative deepening dfs+ strong pruning)