1103 Integer Factorization (30point(s))
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
#include<bits/stdc++.h>
using namespace std;
int n,k,p;
int maxFacSum=0;
int fac[410];
vector<int> ans,temp;
int init(int n,int p){
int i=0,temp=0;
while(temp<=n){
fac[i++]=temp;
temp=pow(i,p);
}
return i;
}
void dfs(int index,int nowK,int sum,int facSum){
if(nowK==k&&sum==n){
if(facSum>maxFacSum){
ans=temp;
maxFacSum=facSum;
}
return;
}
if(sum>n||nowK>k||index==0) return;
temp.push_back(index);
dfs(index,nowK+1,sum+fac[index],facSum+index);
temp.pop_back();
dfs(index-1,nowK,sum,facSum);
}
int main(){
scanf("%d%d%d",&n,&k,&p);
int size=init(n,p);
dfs(size-1,0,0,0);
if(ans.empty()){
printf("Impossible\n");
}else{
printf("%d = ",n);
for(int i=0;i<ans.size();++i){
if(i!=0) printf(" + ");
printf("%d^%d",ans[i],p);
}
printf("\n");
}
}