【PAT】1103 Integer Factorization(30 分)


The KP 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 KP 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 ai​​=bi​​ for i<L and aL​​>bL​​.

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++代碼如下:

 1 #include<iostream>
 2 #include<cmath>
 3 #include<vector>
 4 using namespace std;
 5 
 6 int n, k, p;
 7 vector<int>v;
 8 vector<int>temp,ans;
 9 int sum_g=0;
10 void init() {
11     int t;
12     for (int i = 0; i < 25; i++) {
13         t = pow(i, p);
14         if (t <= n)
15             v.push_back(t);
16         else break;
17     }
18 }
19 void DFS(int index, int sum, int nowk, int sumk) {
20     if (  nowk == k && sum == n) {
21         if (sumk > sum_g) {
22             sum_g = sumk;
23             ans = temp;
24         }
25         return;
26     }
27     if (  sum>n || nowk > k)return;
28     if (index - 1 >= 0) {    
29         temp.push_back(index);
30         DFS(index , sum + v[index], nowk + 1, sumk + index);
31         temp.pop_back();
32         DFS(index - 1, sum, nowk, sumk);
33     }
34 }
35 int main() {
36     cin >> n >> k >> p;
37     init();
38     DFS(v.size() - 1, 0, 0, 0);
39     if (ans.size() > 0) {
40         cout << n << " = ";
41         for (int i = 0; i < ans.size() - 1; i++) 
42             cout << ans[i] << "^" << p << " + ";
43         cout << ans[ans.size() - 1] << "^" << p << endl;
44     }
45     else
46         cout << "Impossible" << endl;
47     return 0;
48 }

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