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 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
#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
vector<int> ans,temp,fac;
int n,k,p,max_fac_sum=-1;
void init(int k,int p){
int i=1;
ans.clear();
temp.clear();
fac.clear();
for(i=0;i<=20;i++){
fac.push_back(pow(i,p));
}
}
int dfs(int index,int cur_k,int sum,int fac_sum){
if(cur_k==k&&sum==n){
if(fac_sum>max_fac_sum)
{
ans=temp;
max_fac_sum=fac_sum;
// for(int i=1;i<ans.size();i++)
// {
// cout<<ans[i]<<" ";
// }
// cout<<endl;
}
return 0;
}
if(sum>n||cur_k>k)
return 0;
if(index>0){
temp.push_back(index);
dfs(index,cur_k+1,sum+fac[index],fac_sum+index);
temp.pop_back();
dfs(index-1,cur_k,sum,fac_sum);
}
return 0;
}
int main(){
scanf("%d %d %d",&n,&k,&p);
init(k,p);
dfs(fac.size()-1,0,0,0);
if(max_fac_sum==-1){
printf("Impossible\n");
}
else{
for(int i=0;i<ans.size();i++){
if(i==0){
printf("%d = %d^%d",n,ans[i],p);
}
else{
printf(" + %d^%d",ans[i],p);
}
}
printf("\n");
}
return 0;
}
题意:
给三个数,npk,把n因式分解,分解成p个数的k次方之和。
要求:
这p个数之和最大,如果存在多种情况,则输出字典序最大的那个。(可以重复选)
难点解析:
字典序:从大到小选择。
求最大和:
if(cur_k==k&&sum==n){
if(fac_sum>max_fac_sum)
{
ans=temp;
max_fac_sum=fac_sum;
寒假第一天