PAT A1103. Integer Factorization (30)

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 = n₁^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, ... a_K } is said to be larger than { b₁, b₂, ... b_K } if there exists 1<=L<=K such that a_i=b_i for i<L and a_L>b_L

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 <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <stack>#include <vector>using namespace std;int n ,k ,p;int lowsum,bestlowsum=-1;vector<int> nowq,bestq,q;int power(int x){int ans=1;for(int i=0;i<p;i++)ans*=x;return ans;}void DFS(int index ,int num ,int sum ,int lowsum){if(num==k&&sum==n){if(lowsum>bestlowsum){bestq=nowq;    bestlowsum=lowsum;}return ;}if(num>k||sum>n) return ;if(index >=1){nowq.push_back(index);DFS(index,num+1,sum+power(index),lowsum+index);nowq.pop_back();DFS(index-1,num,sum,lowsum);}}int main(){scanf("%d%d%d",&n,&k,&p);int m;int f=0,ff=0;while(f<=n){q.push_back(p);f=power(++ff);}      DFS(q.size()-1,0,0,0);  if(bestlowsum==-1)  {  printf("Impossible\n");  }  else   {  printf("%d = %d^%d",n,bestq[0],p);  for(int i=1;i<bestq.size();i++)  {  printf(" + %d^%d",bestq[i],p);  }  printf("\n");  }system("pause");return 0;}