PAT甲级——A1103 Integer Factorization

The K−P factorization of a positive integer N is to write N as the sum of the P-th power of Kpositive 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

#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
int n, k, p, maxFacSum = -1;//maxFacSum用来记录最大底数之和
vector<int>fac, ans, temp;//最大底数不超过n的数，底数最优数序列，临时存放
void DFS(int index, int nowK, int sum, int facSum)
{
    if (sum == n && nowK == k)//统计因素个数
    {
        if (facSum > maxFacSum)//更优的组合
        {
            ans = temp;
            maxFacSum = facSum;
        }
        return;
    }
    if (sum > n || nowK > k)return;//超出限制
    if (index - 1 >= 0)//给出数组小角标的限制
    {
        temp.push_back(index);//记录数据
        DFS(index, nowK + 1, sum + fac[index], facSum + index);//选
        temp.pop_back();//弹出数据
        DFS(index - 1, nowK, sum, facSum);//不选
    }
}
int main()
{
    cin >> n >> k >> p;
    for (int i = 0; pow(i, p) <= n; ++i)
        fac.push_back(pow(i, p));//初始化底数不超过n的因素
    DFS(fac.size() - 1, 0, 0, 0);//为了得到最大的因素数组，从最后一位开始向前搜索
    if (maxFacSum == -1)
        cout << "Impossible" << endl;//没有找到满足的序列
    else 
    {
        cout << n << " = ";
        for (int i = 0; i < ans.size(); i++)
            cout << ans[i] << "^" << p << (i == ans.size() - 1 ? "" : " + ");
    }
    return 0;
}