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 (≤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

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 <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
vector<int> fac, ans, temp;
int n, k, p, maxfacsum = -1;
int power(int x) {
    int ans = 1;
    for(int i = 0; i < p; i++) {
        ans *= x;
    }
    return ans;
}
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 facsum) {
    if(sum == n && nowk == k) {
        if(facsum > maxfacsum) {
            ans = temp;
            maxfacsum = facsum;
        }
        return ;
    }
    if(sum > n || nowk > k) {
        return ;
    }
    if(index - 1 >= 0) {
        temp.push_back(index);
        dfs(index, nowk + 1, sum + fac[index], facsum + index);
        temp.pop_back();
        dfs(index - 1, nowk, sum, facsum);
    }
}
int main()
{
    scanf("%d%d%d", &n, &k, &p);
    init();
    dfs(fac.size() - 1, 0, 0, 0);
    if(maxfacsum == -1) {
        printf("Impossible\n");
    } 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;
}