1103. Integer Factorization

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 = n₁^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 12² + 4² + 2² + 2² + 1², or 11²+ 6² + 2² + 2² + 2², 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, a₂, ... 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

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <deque>
#include <queue>
#include <cstring>
#include <vector>
#include <string>
#include <iomanip>
#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <cmath>
#include <algorithm>
using namespace std;
#define max1 10010
#define RH 1
#define LH -1
#define EH 0
vector<int>ans,tempAns,v;
int n,k,p;
int maxSum=-1;
void init()
{
int temp=0,index=1;
while(temp<=n)
{
v.push_back(temp);
temp=pow(index,p);
index++;
}
}
void DFS(int index,int sum,int funk,int sumk)
{
if(funk==k&&sum==n)
{
if(sumk>maxSum)
{
ans=tempAns;
maxSum=sumk;
}
return;
}
if(funk>k||sum>n)return;
if(index>=1)
{
tempAns.push_back(index);
DFS(index,sum+v[index],funk+1,sumk+index);
tempAns.pop_back();
DFS(index-1,sum,funk,sumk);
}
}
int main() {
cin>>n>>k>>p;
init();
DFS(v.size()-1,0,0,0);
if(maxSum==-1)
{
cout<<"Impossible"<<endl;
}
else
{
cout<<n<<" = ";
for(int i=0;i<ans.size();i++)
{
if(i!=0)cout<<" + ";
cout<<ans[i]<<"^"<<p;
}
cout<<endl;
}
return 0;
}