1103 Integer Factorization (30)

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 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.

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<bits/stdc++.h>
using namespace std;

const int maxn=410;

int factor[maxn];

int n,k,p;

vector<int> ans,temp;

int cnt=0;

bool flag=false;

int maxSum = -1;

void init(){
    int i=0,temp=0;
    while(temp<=n){
        factor[i]=(int)pow(i*1.0,p);
        temp=factor[i];
        cnt=i;
        i++;
    }
}


void DFS(int index,int nowk,int sumW,int sumC){
    if(nowk>k||sumC>n)
            return;
    
    if(nowk==k&&sumC==n){
        
        flag=true;
        
        if(sumW>maxSum){
            maxSum=sumW;
            ans=temp;
        }
    }
    
    
    if(index-1>=0){
        temp.push_back(index);
        DFS(index,nowk+1,sumW+index,sumC+factor[index]);
        temp.pop_back();    
        DFS(index-1,nowk,sumW,sumC);
    
    }

    
        
    
    
}



int main(){
    
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    
    cin>>n>>k>>p;
    
    init();
    
    DFS(cnt,0,0,0);
    
    
    if(!flag){
        cout<<"Impossible
";
        return 0;
    }
    
    
    cout<<n<<" =";
    
    for(int i=0;i<ans.size();i++){
        if(i>0)
            cout<<" +";
        
        cout<<" "<<ans[i]<<"^"<<p;
    }
    
    cout<<endl;
    
    return 0;
    
}