【PTA-A】1103 Integer Factorization (30 分)（DFS）

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

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int n, p, k, maxdi=-1;
vector<int> fac, ans, temp;

int power(int x) {
	int answer = 1;
	for (int i = 0; i < p; i++) {
		answer *= x;
	}
	return answer;
}
void init() {
	int i = 0, temp = 0;
	while (temp <= n) {
		fac.push_back(temp);
		temp = power(++i);
	}
}

void DFS(int index, int nowK, int sum, int di) {
	if (sum == n && nowK == k) {
		if (di > maxdi) {
			ans = temp;
			maxdi = di;
		}
		return;
	}
	if (sum > n || nowK > k)return;
	if (index - 1 >= 0) {
		temp.push_back(index);
		DFS(index, nowK + 1, sum + fac[index], di + index);
		temp.pop_back();
		DFS(index - 1, nowK, sum, di);
	}
}
int main() {
	cin >> n >> k >> p;
	init();
	DFS(fac.size() - 1, 0, 0, 0);
	if (maxdi == -1)cout << "Impossible";
	else {
		printf("%d = %d^%d", n, ans[0], p);
		for (int i = 1; i < ans.size(); i++)
			printf(" + %d^%d", ans[i], p);
	}
	return 0;
}