PAT 甲级 1103. Integer Factorization (30)

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本文介绍了一种针对正整数 N 的 K-P 分解算法，即寻找 K 个正整数的 P 次方之和等于 N 的所有可能组合，并确保组合按非递增顺序排列且具有最大因子总和。

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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;
}