1103. 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<stdio.h>
 #include<string>
 #include<iostream>
 #include<string.h>
 #include<sstream>
 #include<vector>
 #include<map>
 #include<stdlib.h>
 #include<queue>
 #include<math.h>
 #include<set>
 using namespace std;

 int k,p;
 int MAX = -;
 vector<int> re;
 void DFS(vector<int>& vv,int n)
 {
     if(vv.size() == k )
     {
         if(n == )
         {
             int sum = ;
             for(int i =  ;i < k;++i)
                 sum += vv[i];
             if(sum >= MAX) // 需要等号，可使得 sequence { a1, a2, ... aK } is said to be larger than { b1, b2, ... bK } i
             {
                 MAX = sum;
                 re = vv;
             }
         }
         vv.pop_back();
         return;
     }
     int low = vv.size() ==  ?  : vv[vv.size() -];//剪枝 使得只有增序情况
     int m = sqrt(double(n));
     for(int i = low ; i <= m;++i)
     {
         int tmp = pow(double(i),p);
         if(n >= tmp)
         {
             vv.push_back(i);
             DFS(vv,n-tmp);
         }else break;
     }
     if(!vv.empty())
         vv.pop_back();
 }

 int main()
 {
     int n;
     scanf("%d%d%d",&n,&k,&p);
     vector<int> vv;
     DFS(vv, n);
     if(re.empty())
     {
         printf("Impossible\n");
     }
     else
     {
         printf("%d = %d^%d",n,re[re.size()-],p);
         for(int i = re.size() - ;i >= ;--i)
         {
             printf(" + %d^%d",re[i],p);
         }
         printf("\n");
     }
     return ;
 }