PAT(dfs)——1103. Integer Factorization (30)

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本文深入探讨了K-P分解算法，这是一种将正整数N表示为K个正整数的P次幂之和的方法。文章提供了详细的算法实现，包括输入输出规范、样例和关键代码片段，展示了如何通过深度优先搜索(DFS)和剪枝策略来解决这个问题。

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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 + … nK^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

题目大意

dfs+剪枝

题目解析：

这个题目一开始怎么提交都超时，本来以为dfs剪枝写的有问题，借鉴了别人的代码还是超时，后来发现问题出在了init函数，这个地方是要取一个尽可能小的上界的。
修改前：

void init() {
	for(int i=0; i<=n; i++)
		N.push_back(pow(i,p));
}

修改后：

void init() {
	for(int i=0; i<=n; i++){
		int temp=pow(i,p);
		if(temp>n) return;
		N.push_back(temp);
	}
		
}

具体代码：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
#define MAXN 400
vector<int> N,res,A;
int maxsum=-1;
int n,k,p;
void init() {
	for(int i=0; i<=n; i++){
		int temp=pow(i,p);
		if(temp>n) return;
		N.push_back(temp);
	}
		
}
void dfs(int tempsum,int sum,int i,int index) {
	if(i==k){
		if(tempsum==n&&sum>maxsum) {
			maxsum=sum;
			res=A;
		}
	}
	else
		while(index>=1){
			if(tempsum+N[index]<=n){
				A[i]=index;
				dfs(tempsum+N[index],sum+index,i+1,index);
			}
			if(index==1)
				return;
			index--;
		}
}

int main() {
	cin>>n>>k>>p;
	A.resize(k);
	init();
	dfs(0,0,0,N.size()-1);
	if(maxsum==-1) {
		printf("Impossible");
		return 0;
	}
	printf("%d = ",n);
	for(int i=0; i<k; i++) {
		printf("%d^%d",res[i],p);
		if(i!=k-1)
			printf(" + ");
	}
	return 0;
}