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 (≤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 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 ibL​.

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

dfs+剪枝，个数大于k，平方和大于n时直接return 

#include 
#include 
using namespace std;
int n, k, p, maxx;
int m[500], ans[500], temp[500];
bool flag;

int power(int a, int b) { //快速幂
	int t = 1;
	while (b > 0) {
		if (b & 1) {
			t *= a;
		}
		a *= a;
		b >>= 1;
	}
	return t;
}

void dfs(int ind, int cnt, int sum, int order) { //当前数，当前平方和，当前和，当前第几个数
	if (order > k) {
		return;
	}
	if (cnt + m[ind] > n) {
		return;
	}
	if (cnt + m[ind] == n && order == k) {
		sum += ind;
		temp[order] = ind;
		flag = 1;
		if (sum > maxx) {
			maxx = sum;
			for (int i = 1; i <= order; i++) {
				ans[i] = temp[i];
			}
		} else if (sum == maxx) {
			bool f = 0;
			for (int i = 1; i <= k; i++) {
				if (temp[i] < ans[i]) {
					f = 1;
					break;
				}
			}
			if (!f) {
				for (int i = 1; i <= k; i++) {
					ans[i] = temp[i];
				}
			}
		}
	} else {
		sum += ind;
		cnt += m[ind];
		temp[order] = ind;
		for (int i = ind; i >= 1; i--) {
			dfs(i, cnt, sum, order + 1);
		}
	}
}

int main() {
	cin >> n >> k >> p;
	for (int i = 0; i <= sqrt(n); i++) {
		m[i] = power(i, p);
	}
	for (int i = sqrt(n); i >= 1; i--) {
		dfs(i, 0, 0, 1);
	}
	if (!flag) {
		cout << "Impossible";
		return 0;
	}
	cout << n << " = ";
	for (int i = 1; i <= k; i++) {
		printf("%d^%d", ans[i], p);
		if (i != k) {
			printf(" + ");
		}
	}
	return 0;
}