【PAT】1103. Integer Factorization (30)

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本文介绍了一个寻找正整数N的K-P分解的算法，即如何将N表示为K个正整数的P次方之和。通过深度优先搜索加剪枝的方法解决了这一问题，并给出了具体的C++实现代码。

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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 (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

分析：DFS + 剪枝。 http://blog.youkuaiyun.com/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){
	 	//因子的个数达到了k个
		if(n == 0){
			//式子的和刚好与N相等
			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\n");
		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\n",finalFactor[i],P);
	}	
	return 0;
}