1103 Integer Factorization （30 分）

1103 Integer Factorization （30 分）

The K−PK-PK−P factorization of a positive integer NNN is to write NNN as the sum of the PPP-th power of KKK positive integers. You are supposed to write a program to find the K−PK-PK−P factorization of NNN for any positive integers NNN, KKK and PPP.

Input Specification:

Each input file contains one test case which gives in a line the three positive integers NNN (≤400\le 400≤400), KKK (≤N\le N≤N) and PPP (1

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 122+42+22+22+1212^2 + 4^2 + 2^2 + 2^2 + 1^212​2​​+4​2​​+2​2​​+2​2​​+1​2​​, or 112+62+22+22+2211^2 + 6^2 + 2^2 + 2^2 + 2^211​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,a2,⋯,aKa_1, a_2, \cdots , a_Ka​1​​,a​2​​,⋯,a​K​​ } is said to be larger than { b1,b2,⋯,bKb_1, b_2, \cdots , b_Kb​1​​,b​2​​,⋯,b​K​​ } if there exists 1≤L≤K1\le L\le K1≤L≤K such that ai=bia_i=b_ia​i​​=b​i​​ for ibLa_L>b_La​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
#include
#include
#include
using namespace std;
int n,k,p;
int maxsum = -1;
vector a,b,v;
void dfs(int index,int ans,int cnt,int sum){
	if(cnt == k){
		if(ans == n && sum > maxsum){
			b = a;
			maxsum = sum;
		}
		return ;
	}
	while(index >= 1){
		if(ans+v[index] <= n){
			a[cnt] = index;
			dfs(index,ans+v[index],cnt+1,sum+index);
		}
		index--;
	}
}
int main(){
	cin >> n >> k >> p;
	int t=0,index = 1;
	a.resize(k);
	while(t <= n){
		v.push_back(t);
		t = pow(index,p);
		index++;
	}
	dfs(v.size()-1,0,0,0);
	if(maxsum == -1){
		printf("Impossible\n");
		return 0;
	}
	printf("%d = ",n);
	for(int i = 0;i < b.size();i++){
		if(i != 0) printf(" + ");
		printf("%d^%d",b[i],p);
	}
	cout << endl;
	return 0;
}