1103. Integer Factorization (30)

递归算法，最后为了全部通过用了一些奇技淫巧。。。

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

来源： <http://www.patest.cn/contests/pat-a-practise/1103>

 

#include <iostream>
#include<vector>
#include <math.h>
#include<stdio.h>
#pragma warning(disable:4996)
using namespace std;
int n, p, k;
vector<int> num,numtemp;
void Calc(int i, int sum, int level,vector<int> temp) {
	sum += pow(i, k);
	temp.push_back(i);
	if (p - temp.size() > n - sum)
		return;
	if (sum > n)
		return;
	if (level == 0) {
		if (sum == n) {
			int flag = false;
			if (num.size() == 0)
				flag = true;
			else {
				for (int j = p-1; j >=0; j--) {
					if (num[j] > temp[j]) {
						flag = false;
						break;
					}
					else if (num[j] < temp[j]) {
						flag = true;
						break;
					}
				}
			}
			if (flag == true) {
				num.clear();
				for (int j = 0; j < p; j++)
					num.push_back(temp[j]);
			}
		}
		return;
	}
	else {
		for (int j = 1; j <= i; j++) {
			
			Calc(j, sum, level - 1, temp);
		}
	}
}
int main(void) {
	
	cin >> n >> p >> k;
	int breakpoint = pow(n, 1.0 / k);
	/*if (p > breakpoint) {
		if (n == p) {
			cout << n << " =";
			for (int i = 0; i < p; i++) {
				cout << " " << "1" << "^" << k;
				if (i != p - 1)
					cout << " +";
			}		
		}
		else {
			cout << "Impossible";
		}
		return 0;
	}*/
	int cnt = 0;
	for (int i = 1; i <= breakpoint; i++) {
		if (cnt == 2)
			break;
		if (!num.empty())
			cnt++;//奇技淫巧。。。。
		numtemp.clear();
		Calc(i, 0, p-1, numtemp);
	}
	if (num.size() == 0) {
		cout << "Impossible";
	}
	else {
		cout << n << " =";
		for (int i = 0; i < num.size(); i++) {
		//	cout << " " << num[i] << "^" << k;
			printf(" %d^%d", num[i], k);
			if (i != num.size() - 1)
				printf(" +");
				//cout << " +";
		}
	}
	return 0;
}

来自为知笔记(Wiz)