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 <iostream>
#include <cmath>
#include <numeric>
#include <vector>
using namespace std;
int N, K, P, POW[22], optSum; 
vector<int> opt, temp;

void dfs(int start, int val, int sum){
    if(temp.size() > K || val > N || start == 0)    return; 
    if(val == N && temp.size() == K){//满足条件 
        if(opt.size() == 0)    opt = temp, optSum = sum;//第一次得到满足条件的序列 
        else{
            if(sum > optSum)    opt = temp, optSum = sum;
            else if(optSum == sum && temp > opt)    opt = temp;
        }
        return;
    }
    temp.push_back(start);
    dfs(start, val+POW[start], sum + start);//选 
    temp.pop_back();
    dfs(start-1, val, sum); //不选 
}

void print(vector<int> &v){
    printf("%d = ", N);
    for(int i = 0; i < v.size(); ++  i){
        printf("%d^%d", v[i], P);
        if(i != v.size()-1)    printf(" + ");
    }
}

int main()
{
    scanf("%d%d%d", &N, &K, &P);
    int x = sqrt(N)+1;
    for(int i = 1; i <= x; i ++){//打表 
        POW[i] = pow(i, P);
    }
    dfs(x, 0, 0);
    if(opt.size() == 0)    printf("Impossible\n");
    else print(opt);
    return 0;
}