8.1 深度优先搜索：A1103 Integer Factorization

A1103 Integer Factorization

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

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 ib​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 
#include 
#include 
#include 
#include 
#include 
using namespace std;
int n,k,p,maxFacSum=-1;
//fac记录0^p,1^p...，ans存放最优底数序列，temp存放递归中的临时底数序列
vector fac,ans,temp;
int power(int x){
    int ans=1;
    for(int i=0;imaxFacSum){
            ans=temp;
            maxFacSum=facSum;
        }
        return ;
    }
    if(sum>n || nowK>k) return;//这种情况下不会产生答案，直接返回
    if(index-1>=0){
        temp.push_back(index);
        DFS(index,nowK+1,sum+fac[index],facSum+index);
        temp.pop_back();
        DFS(index-1,nowK,sum,facSum);
    }
}
int main(){
    scanf("%d%d%d",&n,&k,&p);
    init();
    DFS(fac.size()-1,0,0,0);
    if(maxFacSum==-1) printf("Impossible\n");
    else{
        printf("%d = %d^%d",n,ans[0],p);
        for(int i=1;i