PAT 1103 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 (≤), K (≤) and P (1). 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 1, or 1, 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 { , } is said to be larger than { , } if there exists 1 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<iostream>
#include<cmath>
#include<vector>
using namespace std;
vector<int> temp, path;
int n, p, k, maxx=-1, t;
bool flag=true;
/*
  dfs的回溯中弹出和压栈的位置很重要
*/
void dfs(int a, int b, int c){
    if(b==0 && c==0){
        int tempMax=0;
        for(int i=0; i<temp.size(); i++) tempMax+=temp[i];
        if(maxx<tempMax){
            maxx = tempMax;
            path = temp;
        }else if(maxx==tempMax && path<temp) path=temp;
        temp.pop_back();
        return;
    }else if(c<=0 || b<=0){
        temp.pop_back();
        return;
    }
    for(int i=a; i>=1; i--){
        if(b-c*pow(i, p)>0) break;  //解决超时的关键点
        if(b-pow(i, p)<0) {
            while(b-pow(i, p)<0 && i>=1) --i;
            temp.push_back(i);
            dfs(i, b-pow(i, p), c-1);
        }else{
            temp.push_back(i);
            dfs(i, b-pow(i, p), c-1);
        }
    }
    temp.pop_back();//弹出的位置在for循环之外，很重要；
}
int main(){
  scanf("%d%d%d", &n, &k, &p);
  t =pow(n, 1.0/p) > n/k+1 ? pow(n, 1.0/p) : (n/k+1);
  dfs(t, n, k);
  if(path.size()==0) cout<<"Impossible"<<endl;
  else {
    printf("%d = ", n);
    for(int i=0; i<path.size(); i++) {
      if(i!=0) printf(" + ");
      printf("%d^%d", path[i], p);
    }
  }
  return 0;
}