PAT 1103. Integer Factorization

最新推荐文章于 2020-07-28 19:41:56 发布

原创最新推荐文章于 2020-07-28 19:41:56 发布 · 349 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种求解正整数 N 的 K-P 分解问题的算法实现，即找到 K 个正整数的 P 次幂之和等于 N 的所有可能组合，并给出了具体的 C++ 实现代码。

DFS算法（可能简单的循环也行），注意跳出循环的条件控制

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

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>

using namespace std;

int N, K, P;
int maxN;
vector<int> t;
vector<vector<int> > rst;

int mypow(int base, int exp)
{
	int rst = 1;
	for(int i=0; i<exp; i++)
		rst *= base;
	return rst;
}

void dfs(int remain, int depth, int start)
{
	if(remain < (K - depth) * mypow(start, P))	return;
	if(depth == K) {
		if(remain == 0) {
			rst.push_back(t);
		}
		return;
	}

	for(int i=start; i<=maxN; i++) {
		t.push_back(i);
		dfs(remain-mypow(i, P), depth+1, i);
		t.pop_back();
	}
}

int main()
{
	cin >> N >> K >> P;
	maxN = (int)sqrt((double)N);

	dfs(N, 0, 1);

	if(!rst.empty()) {
		//cout << rst.size() << endl;
		//for(int j=0; j<rst.size(); j++) {
		int idx = rst.size()-1;
		cout << N << " = " << rst[idx][rst[idx].size()-1] << "^" << P;
			for(int i=rst[idx].size()-2; i>=0; i--)
				cout << " + " << rst[idx][i] << "^" << P;
			//cout << rst[idx][] << endl;
		//}			

	} else {
		cout << "Impossible";
	}
	return 0;
}