PAT甲級——1103 Integer Factorization (DFS)


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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

題目大意:將一個正整數N分解成K個正整數的P次方和,在多個結果里面找出因子之和最大的,若因子之和相同,字典序大的為答案。

思路:主要是DFS的思想,建立一個數組F,用來儲存 1~m的P次方,m^P為≤N的最大正整數。find()里面傳入四個變量,n為當前find()里面的for循環次數;cnt初始值為K,cnt=0作為遞歸的邊界;tmpSum儲存因子之和;sum是總和,sum=N才是符合條件的備選答案~

下一層的遞歸里的n總是小於等於上一層遞歸里的n,所以保證了字典序,不需要畫蛇添足地寫compera函數來篩選答案了(一開始就是因為這個操作導致測試點2答案錯誤),若無必要,勿增操作。

 1 #include <iostream>
 2 #include <vector>
 3 #include <cmath>
 4 using namespace std;
 5 int N, K, P, m, fSum = -1;
 6 vector <int> ans, F, tmpA;
 7 void find(int n,int cnt, int tmpSum, int sum);
 8 
 9 int main()
10 {
11     scanf("%d%d%d", &N, &K, &P);
12     int i = 1;
13     F.push_back(0);
14     while (1) {
15         int x = pow(i, P);
16         if (x > N)
17             break;
18         else {
19             F.push_back(x);
20             i++;
21         }
22     }
23     m = F.size() - 1;
24     find(m, K, 0, 0);
25     if (ans.empty()) {
26         printf("Impossible\n");
27         return 0;
28     }
29     printf("%d =", N);
30     for (int i = 0; i < K; i++) {
31         printf(" %d^%d", ans[i], P);
32         if (i < K - 1) {
33             printf(" +");
34         }
35     }
36     printf("\n");
37     return 0;
38 }
39 void find(int n, int cnt, int tmpSum, int sum) {
40     if(n==0) return;
41     if (cnt == 0) {
42         if (fSum < tmpSum) {
43             if (sum == N) {
44                 ans = tmpA;
45                 fSum = tmpSum;
46             }
47         }
48         return;
49     }
50     for (int i = n; i > 0; i--) {
51         if (sum <= N) {
52             tmpA.push_back(i);
53             find(i, cnt - 1, tmpSum + i, sum + F[i]);
54             tmpA.pop_back();
55         }
56     }
57 }

 


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