PAT甲级真题 1103 Integer Factorization (30分) C++实现（dfs+剪枝+备忘录，经典题目）

最新推荐文章于 2022-02-04 09:59:15 发布

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最新推荐文章于 2022-02-04 09:59:15 发布 · 357 阅读

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#dfs #剪枝 #备忘录 #超时 #因子

该博客详细介绍了如何使用深度优先搜索（DFS）、剪枝和备忘录技术解决PAT甲级竞赛中关于整数因子分解的问题。博主在初始尝试时遇到超时问题，通过不断优化，包括使用幂值表减少计算、剪枝策略以及从大到小的遍历顺序，最终成功实现了AC（Accepted）代码。博客内容包括题目描述、解题思路和优化后的AC代码。

题目

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 ni (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 ibL
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

思路

很棒的一道题，折磨我了整整一下午。

基本思路是遍历+剪枝，写出代码不难，但很容易超时。

由于N<=400，p>=2，所以因子最大值为20，遍历起来复杂度可以接受。

深度优先遍历所有因子，记录当前路径以及节点和，当出现可行路径时，若节点和更大或者节点和相等但序列更大则替换为最优路径。

写了如下代码，果断3个测试点超时。

最初的代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;

vector<vector<int> > allPath;

bool cmp(int a, int b){
   
   
    return a > b;
}

void findFactor(int n, int k, int p, vector<int> path){
   
   
    if (k==1){
   
   
        int j = 1;
        while (pow(j, p) < n) j++;
        if (pow(j, p) == n){
   
   
            path.push_back(j);
            allPath.push_back(path)

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