【PAT】1103. Integer Factorization (30)

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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 (1Output Specification:
For each case, if the solution exists, output in the format:
N = n1^P + ... nK^P
where ni (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, ... aK } is said to be larger than { b1, b2, ... bK } if there exists 1<=L<=K such that ai=bi for ibL
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
분석: DFS + 가지치기. http://blog.csdn.net/xyt8023y/article/details/48465993 
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
int N, K, P;
vector<int> finalFactor;
long sum = 0;

bool dfs(int n, int cur, vector<int> &factor ){
	 if(cur == K){
	 	// k 
		if(n == 0){
			// N 
			int tmpSum = 0;
			for(int i=0; i<factor.size(); i++){
				tmpSum += factor[i];
			} 
			if(tmpSum >= sum){
				finalFactor = factor;	
				sum = tmpSum;		
			}
			return true;			
		}else{
			return false;
		}
	 }	
	 int higher = sqrt(n);
	 int lower = cur>0? factor[cur-1]:1;
	 for(int i=lower; i<=higher; i++){
	 	long tmp = pow(i,P);
		if(tmp > n){
			return false;
		}else{	 					 
			factor[cur] = i;		 
			dfs(n-tmp, cur+1, factor);
		}
	 }	 
	 return true;
}

int main(int argc, char** argv) {
	scanf("%d%d%d",&N,&K,&P);
	vector<int> factor(K);
	dfs(N,0,factor);
	if(finalFactor.size() == 0){
		printf("Impossible
");
		return 0;
	}	 
	printf("%d = ",N);
	for(int i=finalFactor.size()-1; i>=0; i--){
		if(i!=0)
			printf("%d^%d + ",finalFactor[i],P);
		else
			printf("%d^%d
",finalFactor[i],P);
	}	
	return 0;
}
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