PAT甲级——A1103 Integer Factorization

The K−P factorization of a positive integer N is to write N as the sum of the P-th power of Kpositive 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 (≤), K (≤) and P (1). 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 1, or 1, 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 { , } is said to be larger than { , } if there exists 1 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

 #include <iostream>

 #include <vector>

 #include <cmath>

 #include <algorithm>

 using namespace std;

 int n, k, p, maxFacSum = -;//maxFacSum用来记录最大底数之和

 vector<int>fac, ans, temp;//最大底数不超过n的数，底数最优数序列，临时存放

 void DFS(int index, int nowK, int sum, int facSum)

 {

     if (sum == n && nowK == k)//统计因素个数

     {

         if (facSum > maxFacSum)//更优的组合

         {

             ans = temp;

             maxFacSum = facSum;

         }

         return;

     }

     if (sum > n || nowK > k)return;//超出限制

     if (index -  >= )//给出数组小角标的限制

     {

         temp.push_back(index);//记录数据

         DFS(index, nowK + , sum + fac[index], facSum + index);//选

         temp.pop_back();//弹出数据

         DFS(index - , nowK, sum, facSum);//不选

     }

 }

 int main()

 {

     cin >> n >> k >> p;

     for (int i = ; pow(i, p) <= n; ++i)

         fac.push_back(pow(i, p));//初始化底数不超过n的因素

     DFS(fac.size() - , , , );//为了得到最大的因素数组，从最后一位开始向前搜索

     if (maxFacSum == -)

         cout << "Impossible" << endl;//没有找到满足的序列

     else

     {

         cout << n << " = ";

         for (int i = ; i < ans.size(); i++)

             cout << ans[i] << "^" << p << (i == ans.size() -  ? "" : " + ");

     }

     return ;

 }

巴特西