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
将N分解为K个整数的P次幂的和,不唯一时输出和最大的K个数,还不唯一选字典序大的。试了一下,按照题目数据大小,分解出来的可能组合最多有几万种,于是放弃了找到所有序列后再来统计和最大的,而是每找到一个更新一下(当然也可以全部存下来再判断,试过最坏时间多30%,空间消耗也更多)。递归地枚举每一位,需要保证后面的数不大于前面的数,这样还可以省掉很多暴力枚举时间。
#include<stdio.h>
#include<math.h>
#include<vector>
#include<algorithm>
using namespace std;
vector<int> ans;
vector<int> tmp;
int N,K,P;
int summax=0;//当前最大和
bool newans(){
int sum=0;
for(int i=0;i<K;i++){
sum+=tmp[i];
}
if(sum>summax){//有更大的和
summax=sum;
return true;
}else if(sum==summax){//和相等,有更大序列
for(int i=0;i<K;i++){
if(tmp[i]!=ans[i])
return tmp[i]>ans[i];
}
}else{
return false;
}
}
void kpn(int k,int n){
if(k==1){
int t=pow(n,1.0/P);
if((int)pow(t*1.0,P)==n){
tmp.push_back(t);
if(newans()){//有新结果
ans.insert(ans.begin(),tmp.begin(),tmp.end());
}
tmp.pop_back();
}
return;
}
int min=pow(n*1.0/k,1.0/P);
int t=pow(min*1.0,P)*k;
if(t<n)
min++;
int max=pow((n-k+1)*1.0,1.0/P);
if(tmp.size()!=0&&max>tmp.back())//保证后面的数不大于前面的数
max=tmp.back();
for(int i=min;i<=max;i++){//第k位从小到大尝试
tmp.push_back(i);
int nn=n-pow(i*1.0,P);
kpn(k-1,nn);
tmp.pop_back();
}
}
int main(){
scanf("%d%d%d",&N,&K,&P);
kpn(K,N);
if(ans.size()==0){
printf("Impossible\n");
return 0;
}
printf("%d = ",N);
for(int i=0;i<K;i++){
if(i!=0)printf(" + ");
printf("%d^%d",ans[i],P);
}
return 0;
}