【PAT】【Advanced Level】1103. Integer Factorization (30)

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 + ... nK^P

where ni (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

原题链接：
https://www.patest.cn/contests/pat-a-practise/1103

思路：

DFS+回溯+剪枝

DFS搜索因子

剪枝：

最大层数为K

保证搜索序列的是増序序列。

CODE：

#include<iostream>
#include<cmath>
#include<vector>
#include<cstdio>
#include<algorithm>
#define N 410
using namespace std;
typedef struct S
{
int sum;
int num
;
};
vector<S> res;
vector<int> pa;
int nn,kk,pp;
void dfs(int n,int k, int p,int f)
{
if (k!=1)
{
for (int i=f;pow(i,p)<=n/k;i++)
{
int temp=(pow(i,p)+0.5);
pa.push_back(i);
dfs(n-temp,k-1,p,i);
pa.erase(pa.end()-1);
}
}
else
{
int t=-1;
for (int i=1;pow(i,p)<=n;i++)
{
int pp=(pow(i,p)+0.5);
if (pp==n)
{
t=i;
break;
}
}
if (t==-1) return ;
pa.push_back(t);
S ne;
ne.sum=0;
for(int i=0;i<pa.size();i++)
{
ne.sum+=pa[i];
ne.num[i]=pa[i];
}
res.push_back(ne);
pa.erase(pa.end()-1);
}
}
bool cmp(S a,S b)
{
if (a.sum==b.sum)
{
for (int i=0;i<kk;i++)
{
return a.num[i]>b.num[i];
}
}
else
{
return a.sum>b.sum;
}
}
int main()
{
scanf("%d %d %d",&nn,&kk,&pp);
dfs(nn,kk,pp,1);
if (res.size()==0)
{
printf("Impossible");
}
else
{
sort(res.begin(),res.end(),cmp);
S pt=res[0];
printf("%d = ",nn);
for (int i=kk-1;i>=0;i--)
{
printf("%d^%d",pt.num[i],pp);
if (i!=0) printf(" + ");
}
}
return 0;
}