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?Pfactorization 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 (≤), K (≤) and P (1). 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 1, or 1, 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 { , } is said to be larger than { , } if there exists 1 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

DFS問題關鍵：參數的選取問題（要考慮最優解的比較），邊界條件判斷

此處讓node從大到小順序選擇，就可以盡早選中字典序高的

而DFS中第四個參數的選擇是讓多方案判斷時的時間復雜度為O（1）

#include<cstdio>
#include
 
<cmath>
#include<vector>
#include<algorithm>
using namespace std;
vector<int> fac,res,temp;
int n,k,p,maxfac=-1;//記錄最大底數和，判斷最優 
void init(){//預處理fac數組 
    for(int i=0;i<=n;i++){
        if(pow(i,p)>n) break;
        fac.push_back(pow(i,p));
    }
}

void DFS(int node,int sum,int num,int max){
    //節點下標，選數和，選數量， 底數和 
    if(sum==n&&num==k){
        if(max>maxfac) {
            res=temp;
            maxfac=max;
        }
        return;
    }
    else if(node<1||sum>n||num>k) return;
    else{//對於可重復選取問題，先走選分支 
        temp.push_back(node);
        DFS(node,sum+fac[node],num+1,max+node);
        temp.pop_back();
        DFS(node-1,sum,num,max);
    }
}

int main(){
    scanf("%d %d %d",&n,&k,&p);
    init();
    DFS(fac.size()-1,0,0,0);
    if(maxfac==-1) printf("Impossible\n");
    else{
        printf("%d = %d^%d",n,res[0],p);
        for(int i=1;i<res.size();i++)
            printf(" + %d^%d",res[i],p);
    }
    return 0;
}

1103 Integer Factorization （DFS）