1103 Integer Factorization（30 分）(cj)

1103 Integer Factorization（30 分）

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

没想到这题细节很多，有个测试点会超时，那就 位运算求幂 ，保存幂， 从大开始找元素，优化到尽可能小不可能不过的。  

找解就是用dfs求解。 

code

#pragma warning(disable:4996)
#include <iostream>
#include <math.h>
#include <vector>
using namespace std;
vector<int> varr, addvarr, vres, addvres, curres;
int k, n, p, curmin = 0;
void dfs(int pos, int sum, int num);
int main() {
	cin >> n >> k >> p;
	for (int i = 1;; ++i) {
		int tmp = pow(i, p);
		if (tmp > n) break;
		varr.push_back(tmp);
		addvarr.push_back(i);
	}
	dfs(varr.size() - 1, 0, 0);
	if (curres.size() == 0) {
		cout << "Impossible";
	}
	else {
		cout << n << " = ";
		for (int i = 0; i < curres.size(); ++i) {
			if (i != 0) cout << " + ";
			cout << curres[i] << '^' << p;
		}
	}
	cout << endl;
	system("pause");
	return 0;
}
void dfs(int pos, int sum, int num) {
	if (sum > n || num > k) return;
	if (sum == n && num == k) {
		int ssum = 0;
		for (int i = 0; i < addvres.size(); ++i) {
			ssum += addvres[i];
		}
		if (ssum > curmin) {
			curres = addvres;
			curmin = ssum;
		}
		return;
	}
	for (int i = pos; i >= 0; --i) {
		addvres.push_back(i + 1);
		dfs(i, sum + varr[i], num + 1);
		addvres.pop_back();
	}
}