1103 Integer Factorization （30 point(s)）

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

分析：

本題是用到圖的深度優先遍歷演算法。

把從 1 到底數最大值存放 factor 動態陣列中。

從而把問題轉化為，在 factor 數組裡面選 K 個數相加等於N，即多對多關係（圖）。

而找這 K 個數即通過圖的深度優先遍歷來找到最優的底數系列。

詳細程式碼：

#include <iostream>   
#include <algorithm> 
#include <math.h>
#include <vector>
using namespace std;
  
int n,k,p;
int maxfacsum=-1;
vector<int> factor,ans,temp;

static void DFS(int index,int nowk,int sum,int facsum){
    if(sum==n && nowk==k){ // 條件滿足
        if(facsum > maxfacsum){
            ans = temp;
            maxfacsum = facsum;
        }
        return;
    } 
    if(sum>n || nowk>k){   // 不符合要求
        return;
    }

    if(index-1>=0){
        temp.push_back(index);
        DFS(index,nowk+1,sum+factor[index],facsum+index);
        temp.pop_back();
        DFS(index-1,nowk,sum,facsum);
    }
}

// 1103 Integer Factorization （30 point(s)）
int main(void){    
    cin>>n>>k>>p;
     
    int temp=0,j=0;
    while(temp<=n){ //存放 P（2~7） 即底數肯定小於 factor.size()
        factor.push_back(temp);
        temp = pow(++j,p);
    }

    DFS(factor.size()-1,0,0,0); // 深度優先遍歷

    if(maxfacsum == -1){  // 輸出
        cout<<"Impossible"<<endl;
    }else{
        cout<<n<<" = "<<ans[0]<<"^"<<p;
        for(int i=1;i<ans.size();++i){
            cout<<" + "<<ans[i]<<"^"<<p;;
        }
    }

    return 0;
} // jinzheng 2018.9.19 15:33