POJ 3134 Power Calculus (IDA*)

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于 2021-08-05 21:08:24 发布

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本文链接：https://blog.youkuaiyun.com/qq_51392086/article/details/119426634

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该博客探讨了如何通过最少的乘法和除法运算来计算x^n，其中x是初始值，n是正整数。文章介绍了不同的算法策略，包括利用平方和倍增来减少计算步骤，并展示了如何在拥有除法操作时进一步优化计算过程。通过递归和迭代加深搜索的方法，找到达到x^n所需的最小操作数。示例展示了对于不同输入n的解决方案，如n=31、70、91等，输出了对应的最少操作次数。

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题目描述
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.
输入描述:
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.
输出描述:
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.
示例
输入
1
31
70
91
473
512
811
953
0
输出
0
6
8
9
11
9
13
12

题目大意：给出正整数n,若只能使用乘法或除法，输出使x经过运算（自己乘或除自己，以及乘或除运算过程中产生的中间结果）变成x^n的最少步数
若干行数据，对应每行输入的n所需的步数

由于只能乘除已有的数字，一定是x的倍数，等价于从数字1开始，用加减法最少多少次能得到n
初始值都是1，然后递归求解，因为迭代加深搜索（指定递归深度，每次DFS都不超过这个深度），所以时机不对的时候就回来，然后又使用了评估函数进行优化

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iomanip>

using namespace std;

//只看指数即可，从1开始，直到加到n(只允许使用加减)
int depth,a[1005],n;
bool IDA(int pos,int depth)
{
    if(pos>depth) return 0;//IDDFS
    if(a[pos]==n) return 1;
    if(a[pos]<<(depth-pos)<n) return 0;//估价函数，最快的倍增都不能到达，不可实现//剪枝操作
    for(int i=0;i<=pos;i++)
    {
        a[pos+1]=a[pos]+a[i];//加操作
        if(IDA(pos+1,depth)) return 1;
        a[pos+1]=abs(a[pos]-a[i]);//减操作
        if(IDA(pos+1,depth)) return 1;
    }
    return 0;
}

int main()
{
    std::ios::sync_with_stdio(false);
    while(cin>>n&&n)
    {
       for(depth=0;;depth++)
       {
           a[0]=1;
           if(IDA(0,depth)) break;//每次都从0开始，到depth层结束
       }
       cout<<depth<<endl;
    }
	return 0;
}