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

清疚 2022-03-25 16:30 119阅读 0赞

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 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 i<L and aL>bL.

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 <iostream>
    #include <algorithm>
    #include <vector>
    using namespace std;
    int N, K, P, maxFacSum = -1;
    const int MAXN = 100;
    int FAC[MAXN];
    int cindex = 0;
    vector<int> 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;
    }