一、题目
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 (≤400), K (≤N) and P (1<P≤7). 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 122+42+22+22+12, or 112+62+22+22+22, 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 { a1,a2,⋯,aK } is said to be larger than { b1,b2,⋯,bK } if there exists 1≤L≤K 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
二、题目大意
因式分解。
三、考点
DFS
四、注意
1、这个的dfs循环比较特殊,注意使用while和剪枝;
2、参考:https://www.liuchuo.net/archives/2451。
五、代码
#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
using namespace std;
int n, k, p, max_fac_sum=-1;
vector<int> vec_out_path, vec_path, v;
void init() {
int tmp = 0, index = 1;
while (tmp <= n) {
v.push_back(tmp);
tmp = pow(index, p);
index++;
}
}
void dfs(int root, int sum, int k_num, int fac_sum) {
//sum
if (k_num == k) {
if (sum == n && fac_sum > max_fac_sum) {
max_fac_sum = fac_sum;
vec_out_path = vec_path;
}
return;
}
//next dfs
while (root >= 1) {
if (sum + v[root] <= n) {
vec_path.push_back(root);
dfs(root, sum + v[root], k_num + 1, fac_sum + root);
vec_path.pop_back();
}
root--;
}
return;
}
int main() {
//read
cin >> n >> k >> p;
//init
init();
//dfs
dfs(v.size() - 1, 0, 0, 0);
//output
if (max_fac_sum == -1)
cout << "Impossible" << endl;
else {
printf("%d = ", n);
for (int i = 0; i < vec_out_path.size(); ++i) {
if (i != 0)
cout << " + ";
cout << vec_out_path[i] << "^" << p;
}
cout << endl;
}
system("pause");
return 0;
}