1103. Integer Factorization

1103. Integer Factorization (30)

时间限制

1200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 = n₁^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, ... a_K } is said to be larger than { b₁, b₂, ... b_K } if there exists 1<=L<=K 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

没想到什么好办法，就dfs吧，最好自己写一个pow函数，我把对应的p时n的最大值都存起来了，根据p的值决定ni的最大值。

#include<iostream>#include<vector>#include<algorithm>#include<cmath>using namespace std;int n,k,p;int maxn[8] = { 0, 0, 20, 7, 4, 3, 2, 2 };vector<int>t;vector<vector<int>>ans;int ipow(int k,int p){int s=1;while(p--)s=s*k;return s;}bool cmp(vector<int>x,vector<int>y){int xs=0;for(int i=0;i<x.size();++i)xs+=x[i];int ys=0;for(int i=0;i<y.size();++i)ys+=y[i];if(xs!=ys)return xs>ys;elsereturn x>y;}void dfs(int start ,int sum, int count){for (int i = start; i <= maxn[p]; ++i){   int s=sum+ipow(i,p);if(s>n)return;if(count==k&&s<n)continue;if(count==k&&s==n){t.push_back(i);ans.push_back(t);t.pop_back();return;}t.push_back(i);dfs(i,s,count+1);t.pop_back();}}int main(){cin>>n>>k>>p;dfs(1,0,1);if(ans.empty()){cout << "Impossible" << endl;   return 0;  }for (int i = 0; i < ans.size(); ++i){   reverse(ans[i].begin(), ans[i].end());  }   sort(ans.begin(), ans.end(), cmp);cout<<n<<" = "<<ans[0][0]<<"^"<<p;for(int i=1;i<ans[0].size();++i)cout<<" + "<<ans[0][i]<<"^"<<p;//169 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2return 0;}