1103 Integer Factorization (30)

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

思路：

给三个正整数N、K、P，将N表示成K个正整数（可以相同，递减排列）的P次方和，如果有多种方案，选择底数n1+…+nk最大的方案，如果还有多种方案，选择底数序列的字典序最大的方案

C++:

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;

int n,k,p,maxFacSum=-1;
vector<int> fac,ans,temp;
int power(int x){
	int ans=1;
	for(int i=0;i<p;i++){
		ans*=x;
	}
	return ans;
}

void init(){
	int i=0,temp=0;
	while(temp<=n){
		fac.push_back(temp);
		temp=power(++i);
	}
}

void DFS(int index,int nowK,int sum,int facSum){
	if(sum==n&&nowK==k){
		if(facSum>maxFacSum){
			ans=temp;
			maxFacSum=facSum;
		}
		return;
	}
	if(sum>n||nowK>k)return;
	if(index-1>=0){
		temp.push_back(index);
		DFS(index,nowK+1,sum+fac[index],facSum+index);
		temp.pop_back();
		DFS(index-1,nowK,sum,facSum);
	}
} 

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