Integer Factorization

题目描述：

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,⋯,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

思路:

用dfs 遍历所有可能性，并且选择合最大的一组组合

代码:

#include "iostream"
#include "vector"
#include "math.h"
using namespace std;
int n,p,k;
vector<int> res,temp,fac;
int indexSum = -9999;
void dfs(int index, int nowK,int sum, int facSum){
    if(nowK > k || sum >n) return ;
    if (sum == n && nowK == k){
        if(facSum > indexSum ){
            indexSum = facSum;
            res = temp;
        }
        return ;
    }
    if (index >=1){
        temp.push_back(index);
        dfs(index, nowK+1, sum+fac[index], facSum+index);
        temp.pop_back();
        dfs(index-1, nowK, sum , facSum );
    }

}
int main(){
    cin >> n>>k>>p;
    
    int tmp =0;
    int j = 1;
    while(tmp <= n){
        fac.push_back(tmp);
        tmp = int(pow(j++,p));
    }
    dfs(fac.size()-1, 0,0,0);
    for(int i=0;i<res.size(); i++){
        if(i == 0){
            cout << n<<" = ";
            cout << res[i] <<"^"<<p;
        }else
            cout <<" + " << res[i] <<"^"<<p;
    }
    if(indexSum == -9999){
        cout << "Impossible"<<endl;
    }
    return 0;
}