1103 Integer Factorization (30 分)
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 12?2??+4?2??+2?2??+2?2??+1?2??, or 11?2??+6?2??+2?2??+2?2??+2?2??, 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 { a?1??,a?2??,?,a?K?? } is said to be larger than { b?1??,b?2??,?,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
#include<bits/stdc++.h> using namespace std; const int maxn=410; int factor[maxn]; int n,k,p; vector<int> ans,temp; int cnt=0; bool flag=false; int maxSum = -1; void init(){ int i=0,temp=0; while(temp<=n){ factor[i]=(int)pow(i*1.0,p); temp=factor[i]; cnt=i; i++; } } void DFS(int index,int nowk,int sumW,int sumC){ if(nowk>k||sumC>n) return; if(nowk==k&&sumC==n){ flag=true; if(sumW>maxSum){ maxSum=sumW; ans=temp; } } if(index-1>=0){ temp.push_back(index); DFS(index,nowk+1,sumW+index,sumC+factor[index]); temp.pop_back(); DFS(index-1,nowk,sumW,sumC); } } int main(){ ios::sync_with_stdio(false); cin.tie(0); cin>>n>>k>>p; init(); DFS(cnt,0,0,0); if(!flag){ cout<<"Impossible\n"; return 0; } cout<<n<<" ="; for(int i=0;i<ans.size();i++){ if(i>0) cout<<" +"; cout<<" "<<ans[i]<<"^"<<p; } cout<<endl; return 0; }
原文地址:https://www.cnblogs.com/moranzju/p/11332157.html