PAT 1103 Integer Factorization [DFS] [打表]

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本文探讨了 K-P 分解算法，这是一种将正整数 N 表达为 K 个正整数 P 次方之和的方法。文章详细介绍了算法的输入输出规范，提供了示例，并分享了解题过程中的挑战及解决方案。

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

------------------------------------这是题目和解题的分割线------------------------------------

题目挺简单的，就是这个笔记里的思路【DFS】背包问题 | 从N个整数中选择K个数

但是一波三折啊——先是连测试样例都没通过，检查到原来是pow函数有误差，就自己写了个，没问题了。提交，然后又一大堆的段错误和超时，好半天才发现是sqrt函数的问题，sqrt和pow的参数都是double类型，直接用来和整型比较好像会有问题。超时是因为频繁调用自己写的pow函数。最后没辙还是参考了书上的代码。

不要在递归中频繁调用函数，这道题n的数据很小，可以在dfs前提前打表记录到数组中。以及可以让index递减来遍历，这样可以保证字典序大的序列优先被选中，避免比较造成的时间浪费。

#include<cstdio>
#include<vector>

using namespace std;

int n,k,s,maxSum = 0;
vector<int> squ,tmp,out;
 
int powFx(int x)
{
	int sum = 1,i;
	for(i=0;i<s;i++)
		sum *= x;
	return sum;
}

//从0开始不要从1，因为这样数组下标才能一一对应平方和 
void squIni()
{
	int x = 0,i = 0;
	while(x<=n)
	{
		squ.push_back(x);
		x = powFx(++i);
	}
}

void dfs(int index,int sum,int sumSqu,int nowK)
{
	//如果满足条件 
	if(sumSqu==n&&nowK==k)
	{
		//更新和最大值以及最优序列 
		if(sum>maxSum)
		{
			maxSum = sum;
			out = tmp;
		}
		return;
	}
	//中断条件 index最小为1 
	if(sumSqu>n||nowK>k||index<1) return;
	//选择index 
	tmp.push_back(index);
	//index而不是index-1，因为可以重复 
	dfs(index,sum+index,sumSqu+squ[index],nowK+1);
	//不选择index 
	tmp.pop_back();
	dfs(index-1,sum,sumSqu,nowK);
}

int main()
{
	int i;
	scanf("%d%d%d",&n,&k,&s);
	squIni();
	//从squ里的最后一个数字开始寻找 
	dfs(squ.size()-1,0,0,0);
	//如果没有这样的序列 
	if(out.size()==0)
	{
		printf("Impossible");
		return 0;
	}
	printf("%d = ",n);
	for(i=0;i<out.size();i++)
	{
		printf("%d^%d",out[i],s);
		if(i!=out.size()-1)
			printf(" + ");
	}
	return 0;
}