PAT 1103 Integer Factorization（30 分）

最新推荐文章于 2020-03-23 13:00:20 发布

chenxli23

最新推荐文章于 2020-03-23 13:00:20 发布

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本文链接：https://blog.youkuaiyun.com/Hickey_Chen/article/details/81561947

本文介绍了一种针对正整数 N 的 K-P 分解算法，即寻找 N 作为 K 个正整数 P 次幂之和的分解方式。通过深度优先搜索策略，在预计算的幂次方数组中寻找所有可能的组合，并输出具有最大因子和的有效分解方案。

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The K−P factorization of a positive integer N is to write N as the sum of the P-th power of K positive integers. You are supposed to write a program to find the K−P factorization of N for any positive integers N, K and P.

Input Specification:

Each input file contains one test case which gives in a line the three positive integers N (≤400), K (≤N) and P (1<P≤7). The numbers in a line are separated by a space.

Output Specification:

For each case, if the solution exists, output in the format:

N = n[1]^P + ... n[K]^P

where n[i] (i = 1, ..., K) is the i-th factor. All the factors must be printed in non-increasing order.

Note: the solution may not be unique. For example, the 5-2 factorization of 169 has 9 solutions, such as 122+42+22+22+12, or 112+62+22+22+22, or more. You must output the one with the maximum sum of the factors. If there is a tie, the largest factor sequence must be chosen -- sequence { a1,a2,⋯,aK } is said to be larger than { b1,b2,⋯,bK } if there exists 1≤L≤K such that ai=bi for i<L and aL>bL.

If there is no solution, simple output Impossible.

Sample Input 1:

169 5 2

Sample Output 1:

169 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2

Sample Input 2:

169 167 3

Sample Output 2:

Impossible

思路：算法笔记的方法。把不大于的n的i^p的数存到一个数组里，在这个数组里深搜，可以重复使用，所以每次从本身出发深搜，知道搜不到就换另外一个数进行深搜。

程序：

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int n,k,p,maxFacSum = -1;
vector<int> fac,ans,temp;
int power(int x)
{
  int ans = 1;
  for(int i = 0; i < p; i++)
  {
    ans *= x;
  }
  return ans;
}
void init()
{
  int i = 0,temp = 0;
  while(temp <= n)
  {
    fac.push_back(temp);
    temp = power(++i);
  }
}
void dfs(int index,int nowK,int sum,int facSum)
{
  if(sum == n && nowK == k)
  {
    if(facSum > maxFacSum)
    {
      maxFacSum = facSum;
      ans = temp;
    }
    return;
  }
  if(sum > n || nowK > k) return;
  if(index - 1 >= 0)
  {
    temp.push_back(index);
    dfs(index,nowK+1,sum+fac[index],facSum+index);//“选的分支”
    temp.pop_back();
    dfs(index-1,nowK,sum,facSum);//"不选的分支"
  }
}
int main()
{
  scanf("%d%d%d",&n,&k,&p);
  init();
  dfs(fac.size()-1,0,0,0);
  if(maxFacSum == -1)
    printf("Impossible\n");
  else
  {
    printf("%d = %d^%d",n,ans[0],p);
    for(int i = 1; i < ans.size(); i++)
    {
      printf(" + %d^%d",ans[i],p);
    }
  }
  return 0;
}