1103. Integer Factorization (30)

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 + ... 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, ... a_K } is said to be larger than { b₁, b₂, ... b_K } if there exists 1<=L<=K 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<stdio.h>
#include<string>
#include<iostream>
#include<string.h>
#include<sstream>
#include<vector>
#include<map>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<set>
using namespace std;

int k,p;
int MAX = -1;
vector<int> re;
void DFS(vector<int>& vv,int n)
{
    if(vv.size() == k )
    {
        if(n == 0)
        {
            int sum = 0;
            for(int i = 0 ;i < k;++i)
                sum += vv[i];
            if(sum >= MAX) // 需要等号，可使得 sequence { a1, a2, ... aK } is said to be larger than { b1, b2, ... bK } i
            {
                MAX = sum;
                re = vv;
            }
        }
        vv.pop_back();
        return;
    }
    int low = vv.size() == 0 ? 1 : vv[vv.size() -1];//剪枝 使得只有增序情况
    int m = sqrt(double(n));
    for(int i = low ; i <= m;++i)
    {
        int tmp = pow(double(i),p);
        if(n >= tmp)
        {
            vv.push_back(i);
            DFS(vv,n-tmp);
        }else break;
    }
    if(!vv.empty())
        vv.pop_back();
}

int main()
{
    int n;
    scanf("%d%d%d",&n,&k,&p);
    vector<int> vv;
    DFS(vv, n);
    if(re.empty())
    {
        printf("Impossible\n");
    }
    else
    {
        printf("%d = %d^%d",n,re[re.size()-1],p);
        for(int i = re.size() - 2;i >= 0;--i)
        {
            printf(" + %d^%d",re[i],p);
        }
        printf("\n");
    }
    return 0;
}