(PAT 1103) Integer Factorization (深度优先遍历解决背包问题)

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

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 ib​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

解题思路:

1.先把可能组成N的所有数的平方和保存到一个数组中

2.利用DFS遍历数组,对于每个数,有选与不选两种情况

选(可重复):继续重复选取该数,直到不可选取为止

不选:继续遍历下一个数

#include 
#include 
#include 
using namespace std;
int N, K, P, maxFacSum = -1;
const int MAXN = 100;
int FAC[MAXN];
int cindex = 0;
vector temp, ans;
int mpow(int number, int p) {
	int ans = 1;
	for (int i = 1; i <= p; ++i) {
		ans *= number;
	}
	return ans;
}
void initFac() {
	int ntemp = 0;
	while (ntemp <= N) {
		FAC[cindex] = mpow(cindex + 1, P);
		ntemp = FAC[cindex];
		cindex++;
	}
}
void FDFS(int index, int numK, int sum, int facSum) {
	if (sum == N && numK == K) {
		if (facSum > maxFacSum) {
			ans = temp;  //更新ans
			maxFacSum = facSum;
		}
		return;
	}
	if (numK > K || sum > N) {
		return;   //边界条件
	}
	if (index >= 0) {
		temp.push_back(index + 1);
		FDFS(index, numK + 1, sum + FAC[index], facSum + index); //选
		temp.pop_back();
		FDFS(index - 1, numK, sum, facSum);  //不选
	}
}
int main() {
	scanf("%d %d %d", &N, &K, &P);
	initFac();
	FDFS(cindex - 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;
}