1103Integer Factorization(30分)

TheK−Pfactorization of a positive integerNis to writeNas the sum of theP-th power ofKpositive integers. You are supposed to write a program to find theK−Pfactorization ofNfor any positive integersN,KandP.

Input Specification:

Each input file contains one test case which gives in a line the three positive integersN(≤),K(≤) andP(1). 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

wheren[i](i= 1, ...,K) is thei-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 as1, or1, 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 {,} is said to belargerthan {,} if there exists1such thata​i​​=b​i​​fori<Landa​L​​>b​L​​.

If there is no solution, simple outputImpossible.

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：

暴力求解100%會超時，這裡使用DFS進行求解，由題目要求，我們從最大的因子開始從大往小搜，在因子合相同時總是能先搜到最優解然後返回。

Solution(Golang):

package main

import (
	"fmt"
)

var (
	temp, fac []int
	ans       [100]int
	maxsum    int = 0
	n, p, k   int
	flag      = false
)

func pow(n, p int) int {
	a := n
	for i := 1; i < p; i++ {
		a *= n
	}
	return a
}

func init1() {
	fac = append(fac, 0)
	for i := 1; pow(i, p) <= n; i++ {
		fac = append(fac, pow(i, p))
	}
}

func dfs(index, num, value, sumfac int) {
	if num > k || value > n {
		return
	}
	if num == k {
		if value == n && sumfac > maxsum {
			flag = true
			maxsum = sumfac
			for i, v := range temp {
				ans[i] = v
			}

		}
		return
	}
	temp = append(temp, index)
	//fmt.Println(index)
	dfs(index, num+1, value+fac[index], sumfac+index)
	temp = temp[0 : len(temp)-1]
	if index > 1 {
		dfs(index-1, num, value, sumfac)
	}
}

func main() {
	fmt.Scan(&n, &k, &p)
	init1()

	dfs(len(fac)-1, 0, 0, 0)
	//fmt.Println(ans)
	if !flag {
		fmt.Println("Impossible")
	} else {
		fmt.Printf("%d = %d^%d", n, ans[0], p)
		for i := 1; i < k; i++ {
			fmt.Printf(" + %d^%d", ans[i], p)
		}
	}
	return
}

　　

#include<bits/stdc++.h>
usingnamespacestd;
constintmaxn=10010; intn,x; stringa,b; structnode{ intleft,right; };
nodetree[20];
voidinvert(introot){ if(root==-1)return; invert(tree[root].left); invert(tree[root].right); swap(tree[root].left,tree[root].right); } vector<int>ans; voidlayer(introot){ queue<int>q; q.push(root); while(!q.empty()){ intt=q.front(); ans.push_back(t); q.pop(); if(tree[t].left!=-1)q.push(tree[t].left); if(tree[t].right!=-1)q.push(tree[t].right); } }
voidinorder(introot){ if(root==-1)return; inorder(tree[root].left); ans.push_back(root); inorder(tree[root].right); }
voidprintAns(){ for(inti=0;i<ans.size();++i){ cout<<ans[i]; if(i<ans.size()-1){ cout<<""; }else{ cout<<endl; } } }
intmain(){ cin>>n; boolisf[maxn]; fill(isf,isf+n,true); for(inti=0;i<n;++i){ cin>>a>>b; if(a[0]=='-'){ tree[i].left=-1; }else{ x=atoi(a.data()); tree[i].left=x; isf[x]=false; } if(b[0]=='-'){ tree[i].right=-1; }else{ x=atoi(b.data()); tree[i].right=x; isf[x]=false; } } for(inti=0;i<n;++i){ if(isf[i]){ x=i; break; } } invert(x); layer(x); printAns(); ans.clear(); inorder(x); printAns();
return0; }