【PAT】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 (≤), 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 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

C++代码如下：

 #include<iostream>
 #include<cmath>
 #include<vector>
 using namespace std;

 int n, k, p;
 vector<int>v;
 vector<int>temp,ans;
 int sum_g=;
 void init() {
     int t;
     for (int i = ; i < ; i++) {
         t = pow(i, p);
         if (t <= n)
             v.push_back(t);
         else break;
     }
 }
 void DFS(int index, int sum, int nowk, int sumk) {
     if (  nowk == k && sum == n) {
         if (sumk > sum_g) {
             sum_g = sumk;
             ans = temp;
         }
         return;
     }
     if (  sum>n || nowk > k)return;
     if (index -  >= ) {
         temp.push_back(index);
         DFS(index , sum + v[index], nowk + , sumk + index);
         temp.pop_back();
         DFS(index - , sum, nowk, sumk);
     }
 }
 int main() {
     cin >> n >> k >> p;
     init();
     DFS(v.size() - , , , );
     if (ans.size() > ) {
         cout << n << " = ";
         for (int i = ; i < ans.size() - ; i++)
             cout << ans[i] << "^" << p << " + ";
         cout << ans[ans.size() - ] << "^" << p << endl;
     }
     else
         cout << "Impossible" << endl;
     return ;
 }