• 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−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

    Source

    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

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 using namespace std;
     5 int num;
     6 int way[1006];
     7 bool dfs(int n,int step){
     8     if(num>step) return false;
     9     if(way[num]==n) return true;
    10     if(way[num]<<(step-num)<n) return false;//强剪枝
    11     for(int i=0;i<=num;i++){
    12         num++;
    13         way[num]=way[num-1]+way[i];
    14         if(way[num]<=1000 && dfs(n,step)) return true;
    15 
    16         way[num]=way[num-1]-way[i];
    17         if(way[num]>0 && dfs(n,step)) return true;
    18         num--;
    19     }
    20     return false;
    21 }
    22 int main()
    23 {
    24     int n;
    25     while(scanf("%d",&n)==1){
    26         if(n==0){
    27             break;
    28         }
    29 
    30         //迭代加深dfs
    31         int i;
    32         for(i=0;;i++){
    33             way[num=0]=1;
    34             if(dfs(n,i))
    35                 break;
    36         }
    37         printf("%d
    ",i);
    38         
    39     }
    40     return 0;
    41 }
    View Code
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  • 原文地址:https://www.cnblogs.com/UniqueColor/p/4776943.html
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