PAT 甲级 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 <algorithm>#include <cstdio>#include <iostream>#include <vector>#include <queue>using namespace std;vector<int> v, ans, tempAns;int n, k, p, maxFacSum = -1;void init() {int temp = 0, index = 1;while (temp <= n) {v.push_back(temp);temp = pow(index, p);index++;}}void dfs(int index, int tempSum, int tempK, int facSum) {if (tempSum == n&&tempK == k) {if (facSum > maxFacSum) {ans = tempAns;maxFacSum = facSum;}return;}if (tempSum > n || tempK > k) return;if (index >= 1) {tempAns.push_back(index);dfs(index, tempSum + v[index], tempK + 1, facSum + index);tempAns.pop_back();dfs(index - 1, tempSum, tempK, facSum);}}int main() {scanf("%d%d%d", &n, &k, &p);init();dfs(v.size() - 1, 0, 0, 0);if (maxFacSum == -1) {printf("Impossible");return 0;}printf("%d = ", n);for (int i = 0; i < ans.size(); i++) {if (i != 0) printf(" + ");printf("%d^%d", ans[i], p);}return 0;}