PAT 1103 Integer Factorization

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最新推荐文章于 2020-07-28 19:41:56 发布

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本文介绍了一个整数分解问题的解决方案，即如何将一个正整数N表示为K个正整数P次幂之和的形式。文章详细阐述了该问题的输入输出规范，并提供了一个具体的实现算法，包括递归搜索策略来寻找所有可能的分解方案。

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1103. Integer Factorization (30)

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 <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
using namespace std;
int re[21];
int s[400],t[400];
int n,k,p;
int flag,Max=-1;
void Print(){
	printf("%d",n);
	for(int i=0;i<k;i++){
		printf(" %c %d^%d",i==0?'=':'+',t[i],p);
	}
	cout<<endl;
}
void dfs(int loc,int sum,int sumFactor,int step){
	if(sum>n){
		return;
	}
	if(step==k){
		if(sum<n||sumFactor<Max){
			return;
		}
		if(sumFactor==Max){
			for(int i=0;i<k;i++){
				if(s[i]<t[i]){
					return;
				}
			}
		}
		flag=1;
		Max=sumFactor;
		for(int i=0;i<k;i++){
			t[i]=s[i];
		}
		return;	
	}
	for(int i=loc;i>=1;i--){
		s[step]=i;
		dfs(i,sum+re[i],sumFactor+i,step+1);
	}
}
int main(){
	cin>>n>>k>>p;
	for(int i=1;i<=20;i++){
		re[i]=(int)pow(i,p);
	}
	for(int i=20;i>=1;i--){
		s[0]=i;
		dfs(i,re[i],i,1);
	}
	if(!flag){
		cout<<"Impossible"<<endl;
	}
	else{
		Print();
	}
	return 0;
}