PAT甲級1103題解

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 { 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

這道題就是dfs加剪枝，思路還是比較明確的；

#include<cmath>
#include <iostream>
#include<vector>
using namespace std;
int n, k,p,now=0,sum=0;//nkp如題，now存放當前滿足條件的解的最大和，sum是當前解的和
bool flag=false;//判斷是否存在解
int a[410];
vector<int> ans;//臨時存放解數組
vector<int> fans;//滿足條件的數組，最後結果保存在這個數組當中
void print() {//打印最後結果
	cout << n << " =";
	for (int i = 0; i < k; i++) {
		printf("%s%d^%d", i == 0 ? " " : " + ", fans[i], p);
	}cout << endl;
}
void dfs(int temp,int num,int k) {	
	for (int i = temp; i > 0; i--) {
		if (a[i] == num && k == 1) {
			flag = true;
			ans.push_back(i);
			sum += i;
			if (now < sum) {	
				now = sum;
				fans = ans;
				//print();
			}
			ans.pop_back();
			sum -= i;
			continue;
		}
		if (a[i] * k < num)return;
		else if (a[i] <= num) {
			ans.push_back(i);
			sum += i;
			dfs(i, num-a[i], k - 1);
			ans.pop_back();
			sum -= i;
		}
	}
}
int main()
{
	cin >> n >> k >> p;
	/* //t這個值需要動態生成，用後面寫的while去做
	//如果直接給定t的值或者開方去求的話，有一個測試點過不了
	//但是那個測試點可以通過固定t的值等於8或者9，這樣那個測試點就可以通過
	//這個地方到底是因爲什麼原因，目前還沒有搞清楚
	int t = 13; 
	for (int i = 0; i <= t; i++) {
		a[i]=pow(1.0*i, p);
	}*/
	int t = 0, temp = 0;
	while (temp <= n  ) {
		a[t++] = pow(1.0*t, p);
	}

	dfs(t-1,n,k);
	if (!flag) {
		cout << "Impossible" << endl;
		return 0;
	}
	print();
	return 0;
}