题目122：找出计算幂的最高效方法

欧拉计划

Efficient exponentiation

The most naive way of computing n¹⁵ requires fourteen multiplications:

n × n × ... × n = n¹⁵

But using a "binary" method you can compute it in six multiplications:

n × n = n²
n² × n² = n⁴
n⁴ × n⁴ = n⁸
n⁸ × n⁴ = n¹²
n¹² × n² = n¹⁴
n¹⁴ × n = n¹⁵

However it is yet possible to compute it in only five multiplications:

n × n = n²
n² × n = n³
n³ × n³ = n⁶
n⁶ × n⁶ = n¹²
n¹² × n³ = n¹⁵

We shall define m(k) to be the minimum number of multiplications to compute n^k; for example m(15) = 5.

For 1 ≤ k ≤ 200, find ∑ m(k).

题目：

计算 n¹⁵ 的最朴素的算法要 14 次乘法

n × n × ... × n = n¹⁵

如果采用分治的方法，你可以只要六次乘法就完成这一工作：

n × n = n²
n² × n² = n⁴
n⁴ × n⁴ = n⁸
n⁸ × n⁴ = n¹²
n¹² × n² = n¹⁴
n¹⁴ × n = n¹⁵

然而，只用 5 次乘法的方法也是存在的：

n × n = n²
n² × n = n³
n³ × n³ = n⁶
n⁶ × n⁶ = n¹²
n¹² × n³ = n¹⁵

我们定义 m(k) 为计算 n^k 所需的乘法的最少次数，比如 m(15)=5。

对于 1 ≤ k ≤ 200, 求 m(k) 的和。

jerryxjr1220

解答：

#coding:utf-8
L = 201
ways = [[[0]*L] for i in range(L)]
ways[2] = [[2,1]]
for i in range(2,L):
    for eachway in ways[i]:
        for p in eachway:
            n = i + p
            if n < L:
                if min([len(each) for each in ways[n]]) > min([len(each) for each in ways[i]]) + 1:
                    ways[n] = [[n] + eachway]
                elif min([len(each) for each in ways[n]]) == min([len(each) for each in ways[i]]) + 1:
                    ways[n] += [[n] + eachway]

print (sum([(min([len(each) for each in ways[x]])-1) for x in range(2,201)]))

答案：
1582

guosl

//应用带路径的广搜
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <cstring>
using namespace std;

const int nN = 200;
int m[nN + 1];//记录最短路径长度

queue<vector<int> > q;//带路径的广搜队列

int main(void)
{
  for (int i = 1; i <= nN; ++i)
    m[i] = i - 1;
  vector<int> v;//记录搜索的路径
  v.push_back(1);
  q.push(v);
  while (!q.empty())
  {
    vector<int> v0 = q.front();
    q.pop();
    for (int i = 0; i < (int)v0.size(); ++i)//扩展下一个节点
    {
      int k = v0.back();
      k += v0[i];
      if (k <= nN)//检查数据的有效性
      {
        if (v0.size() <= m[k])//找到更短的路径
        {
          m[k] = v0.size();
          v = v0;
          v.push_back(k);
          q.push(v);
        }
      }
    }
  }
  int nS = 0;
  for (int i = 1; i <= nN; ++i)
  {
    cout << i << ": " << m[i] << endl;
    nS += m[i];
  }
  cout << nS << endl;
  return 0;
}

答案：1582