PAT A1103 Integer Factorization (30分)

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805364711604224

题目描述
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.

输入
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.

输出
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 112​​ +62​​ +22​​ +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​​ ,⋯,aK​​ } 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.

样例输入
169 5 2

样例输出
169 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2

代码

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
int n, k, p, maxFacSum = -1;
vector<int> v, ans, tempAns;
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 (tempK == k) {
        if (tempSum == n && facSum > maxFacSum) {
                ans = tempAns;
                maxFacSum = facSum;
        }
        return;
    }
    while(index >= 1) {
        if (tempSum + v[index] <= n) {
            tempAns[tempK] = index;
            dfs(index, tempSum + v[index], tempK + 1, facSum + index);
        }
        if (index == 1) return;
        index--;
    }
}
int main() {
    scanf("%d%d%d", &n, &k, &p);
    init(); 
    tempAns.resize(k);
    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;
}