PAT 1103. Integer Factorization (30)

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本文介绍了一种使用DFS加剪枝技术解决整数分解问题的算法，旨在找到将正整数N表示为K个正整数的P次幂之和的方法。通过预先计算各数的幂值并存储，避免了重复计算，有效提高了算法效率。

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1103. Integer Factorization (30)

时间限制

1200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 = n₁^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, ... 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

这道题我用DFS＋剪枝来做，主要要解决的问题还是超时问题。我们可以事先把各个数的各个次方求出来并保存在数组中，这样一来就不用在后面的DFS中反复计算，以节约时间。并且根据题意分析，N<=400, 1<P<=7，所以实际需要事先求的数并不多，用pow()函数可以解决，不会超时。

代码如下：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
int N,K,P;
int fact[21][8]={-1};
int limit[21] = {0,8,8,6,5,4, 4,4,3,3,3, 3,3,3,3,3, 3,3,3,3,3};
vector<int> tmpAnswer;
vector<vector<int> > allAnswer;
void dfs(int startPos, int sum, int count){
	for(int i = startPos; fact[i][P] > 0; i++){
		int sumTmp = sum + fact[i][P];
		if(sumTmp > N)
			return;
		if(count == K && sumTmp < N)
			continue;
		if(count == K && sumTmp == N){
			tmpAnswer.push_back(i);
			allAnswer.push_back(tmpAnswer);
			tmpAnswer.pop_back();
			return;
		}
		tmpAnswer.push_back(i);
		dfs(i,sumTmp,count+1);
		tmpAnswer.pop_back();
	}
}
int cmp(const vector<int> a, const vector<int> b){
	int sumA = 0, sumB = 0;
	for(int i = 0; i < a.size(); i++){
		sumA += a[i];
		sumB += b[i];
	}
	if(sumA != sumB)
		return sumA > sumB;
	return a > b;
}
int main(void)
{
	for(int i = 1; i < 21; i++)
		for(int j = 2; j < limit[i]; j++){
			int tmp = pow(i,j);
			fact[i][j] = tmp;
		}
	cin>>N>>K>>P;
	dfs(1,0,1);
	if(allAnswer.size() <= 0){
		cout<<"Impossible"<<endl;
		return 0;
	}
	for(int i = 0 ; i < allAnswer.size(); i++)
		reverse(allAnswer[i].begin(),allAnswer[i].end());
	sort(allAnswer.begin(),allAnswer.end(),cmp);
	cout<<N<<" = "<<allAnswer[0][0]<<"^"<<P;
	for(int i = 1; i < allAnswer[0].size(); i++){
		cout<<" + "<<allAnswer[0][i]<<"^"<<P;
	}
	return 0;
}