PAT甲级——A1103 Integer Factorization

The

Input Specification:

Each input file contains one test case which gives in a line the three positive integers

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 $1, or 1, or more. You must output the one with the maximum sum of the factors. If there is a tie, the largest factor sequence must be chosen -- sequence {, } is said to be larger than {, } if there exists 1 such that$

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

 1 #include <iostream>
 2 #include <vector>
 3 #include <cmath>
 4 #include <algorithm>
 5 using namespace std;
 6 int n, k, p, maxFacSum = -1;//maxFacSum用来记录最大底数之和
 7 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
 8 void DFS(int index, int nowK, int sum, int facSum)
 9 {
10     if (sum == n && nowK == k)//统计因素个数
11     {
12         if (facSum > maxFacSum)//更优的组合
13         {
14             ans = temp;
15             maxFacSum = facSum;
16         }
17         return;
18     }
19     if (sum > n || nowK > k)return;//超出限制
20     if (index - 1 >= 0)//给出数组小角标的限制
21     {
22         temp.push_back(index);//记录数据
23         DFS(index, nowK + 1, sum + fac[index], facSum + index);//选
24         temp.pop_back();//弹出数据
25         DFS(index - 1, nowK, sum, facSum);//不选
26     }
27 }
28 int main()
29 {
30     cin >> n >> k >> p;
31     for (int i = 0; pow(i, p) <= n; ++i)
32         fac.push_back(pow(i, p));//初始化底数不超过n的因素
33     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
34     if (maxFacSum == -1)
35         cout << "Impossible" << endl;//没有找到满足的序列
36     else 
37     {
38         cout << n << " = ";
39         for (int i = 0; i < ans.size(); i++)
40             cout << ans[i] << "^" << p << (i == ans.size() - 1 ? "" : " + ");
41     }
42     return 0;
43 }