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".

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<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
int n,k,p;
int maxfacsum = -1;
  
  //fac存储的是0^p,1^p……不超过n的数，用来加给facsum和n比较
  //ans是最优底数和（最大），temp是临时底数数组，和最优比较 
vector<int> fac,ans,temp;

int power(int x){
    int ans = 1;
    for(int i = 0; i < p; i++){
        ans *= x;
    }
    return ans;
}
void init(){  //将小于n的p次方数依次存在fac数组中 
    int temp = 0,i = 0;
    while(temp <= n){
         fac.push_back(temp);
         temp = power(++i);
    } 
    
}

//index表示访问的fac[index]数，nowK表示实际K是数相加
//facsum 表示当前选择底数之和，sum表示当前选中数之和 
void DFS(int index,int nowK,int facsum,int sum) {
    if(sum == n && nowK == k){
        if(facsum > maxfacsum ){
            ans = temp;
            maxfacsum = facsum;
        }
        return;
    }
    if(sum > n || nowK > k) return;
    if(index - 1 >= 0){
        temp.push_back(index);
        DFS(index,nowK+1,facsum+index,sum+fac[index]);
        temp.pop_back();
        DFS(index - 1,nowK,facsum,sum);
    }
}

int main(){
   
    scanf("%d%d%d",&n,&k,&p);
    init();
    DFS(fac.size()-1,0,0,0);
    if(maxfacsum == -1) printf("Impossible
");
    else{
        printf("%d = %d^%d",n,ans[0],p);
     for(int i = 1; i < ans.size(); i++){
        printf(" + %d^%d",ans[i],p);
      }
    }
    return 0;
}