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".

169 5 2Sample Output 1:
169 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2Sample Input 2:
169 167 3Sample 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;}