1103 Integer Factorization (30 分) dfs 剪枝

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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 = 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 12​2​​+4​2​​+2​2​​+2​2​​+1​2​​, or 11​2​​+6​2​​+2​2​​+2​2​​+2​2​​, 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 { a​1​​,a​2​​,⋯,a​K​​ } is said to be larger than { b​1​​,b​2​​,⋯,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

这道题题意很简单，给出s,n,k找一组数满足s=n个数的k次方之和

如果有多组数满足，则取底数之和最大的，且输出的时候底数从大到小输出。

深搜里记录：x-当前值，nown-当前共有n个数，nowsum-当前和

当nown==n&&nowsum==s时满足条件，比较sum和res_sum，如果更大，则替换结果res=v;

#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <queue>
#include <math.h>
#include <algorithm>
#include <iostream>
#define INF 0x3f3f3f3f
using namespace std;
int s,n,k,res_sum;
vector<int> v,res;
void dfs(int x,int nown,int nowsum)
{if(x<=0||nown>n||nowsum>s)return;//printf("%d %d %d\n",x,nown,nowsum);if(nown==n&&nowsum==s){int sum=0;for(int i=0;i<v.size();i++)sum+=v[i];//printf("%d\n",sum);if(sum>res_sum){res_sum=sum;res=v;}return;}v.push_back(x);dfs(x,nown+1,nowsum+pow(x,k));v.pop_back();dfs(x-1,nown,nowsum);
}
int main()
{scanf("%d%d%d",&s,&n,&k);res_sum=-1;dfs(sqrt(s),0,0);if(res_sum!=-1){printf("%d = ",s);sort(res.begin(),res.end());for(int i=res.size()-1;i>0;i--)printf("%d^%d + ",res[i],k);printf("%d^%d\n",res[0],k);}	elseprintf("Impossible\n");
}

 

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