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 thei-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 { a₁,a​₂,⋯,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

Code:

#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int n,k,p,maxFacsum = -1;
vector<int> fac,ans,temp;
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;
}