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 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 i<L and aL>bL.
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<stdio.h>
#include<math.h>
#include<algorithm>
#include<vector> //用vector方便更新
using namespace std;
int seq[500];
int num;
int p;
int k;
int i=0;
vector<int> temp;
vector<int>result;
int maxsum=-1;
void find(int index,int nowk,int sum,int seqsum)
{
if(sum==num&&nowk==k)
{
if(seqsum>maxsum)
{
maxsum=seqsum;
result=temp;
}
return;
}
if(sum>num||nowk>k||index<1)
return;
temp.push_back(index);
find(index,nowk+1,sum+seq[index],seqsum+index);
temp.pop_back();
find(index-1,nowk,sum,seqsum);
}
int main()
{
scanf("%d",&num);
scanf("%d",&k);
scanf("%d",&p);
while(pow(i,p)<=num)
{
seq[i]=pow(i,p);
i++;
}
find(i-1,0,0,0);
int tempk=0;
if(maxsum==-1)
{
printf("Impossible");
}
else
{
while(result.size()!=tempk)
{
if(tempk==0)
{
printf("%d = %d^%d",num,result[tempk],p);
tempk++;
}
else
{
printf(" + %d^%d",result[tempk],p);
tempk++;
}
}
}
}