1103 Integer Factorization（30 分）

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 12?2??+4?2??+2?2??+2?2??+1?2??, or 11?2??+6?2??+2?2??+2?2??+2?2??, 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 { a?1??,a?2??,?,a?K?? } is said to be larger than { b?1??,b?2??,?,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

題目大意：輸入一個數N，以及數量K，指數P，將N用K個數的N次方的和進行表示。如果相同輸出因子數最小的，如果還相同，那麽輸出較大的那個。

//感覺好難，是不是dfs什麽的？果然大佬覺得有趣的題目都這麽難。

PAT 1103 Integer Factorization[難]