【PAT】1103 Integer Factorization（30 分）

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 (≤), K (≤) and P (1). 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 1, or 1, 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 { , } is said to be larger than { , } if there exists 1 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

 

C++代码如下：

#include<iostream>
#include<cmath>
#include<vector>
using namespace std;

int n, k, p;
vector<int>v;
vector<int>temp,ans;
int sum_g=0;
void init() {
    int t;
    for (int i = 0; i < 25; i++) {
        t = pow(i, p);
        if (t <= n)
            v.push_back(t);
        else break;
    }
}
void DFS(int index, int sum, int nowk, int sumk) {
    if (  nowk == k && sum == n) {
        if (sumk > sum_g) {
            sum_g = sumk;
            ans = temp;
        }
        return;
    }
    if (  sum>n || nowk > k)return;
    if (index - 1 >= 0) {    
        temp.push_back(index);
        DFS(index , sum + v[index], nowk + 1, sumk + index);
        temp.pop_back();
        DFS(index - 1, sum, nowk, sumk);
    }
}
int main() {
    cin >> n >> k >> p;
    init();
    DFS(v.size() - 1, 0, 0, 0);
    if (ans.size() > 0) {
        cout << n << " = ";
        for (int i = 0; i < ans.size() - 1; i++) 
            cout << ans[i] << "^" << p << " + ";
        cout << ans[ans.size() - 1] << "^" << p << endl;
    }
    else
        cout << "Impossible" << endl;
    return 0;
}