1103 Integer Factorization (30分) 【DFS】

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
           

題解：

做的第二道DFS題，還是沒能獨立寫出來。需要在認真總結了解下DFS。

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

int n, k, p, maxFacSum = -1;
vector<int> fac, ans, temp;

void init() {
	int i = 1, temp = 0;
	while (temp <= n)
	{
		fac.push_back(temp);
		temp = pow(i, p);
		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() {
	cin >> n >> k >> p;
	init();
	DFS(fac.size() - 1, 0, 0, 0);
	if (maxFacSum == -1) cout << "Impossible";
	else
	{
		cout << n << " = " << ans[0] << '^' << p;
		for (int i = 1; i < ans.size(); i++)
			cout << " + " << ans[i] << '^' << p;
	}
    return 0;
}