PAT-甲级-1103.IntegerFactorization(30)（dfs回溯+减枝）

题目描述：

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 = 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 belarger than { b₁, b₂, ... b_K } if there exists 1<=L<=K such that a_i=b_i for iL>b_L

If there is no solution, simple output "Impossible".

169 5 2

169 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2

169 167 3

Impossible

题目思路：

递归搜索出所有的情况（注意减枝），输出数字和最大的一组即可。

题目代码：

#include 
#include 
#include 
using namespace std;
int n, k, p;
int ans[405];
int sum = 0,cnt = 0;
vectorv[100000];  
void dfs(int cur, int num){
	// 退出条件 
	if(sum > n || cur > k) return ;	
	if(sum == n){  
		if(cur == k){ // 符合条件的结果			
			for(int i = 0; i =1 ;i--){
		int temp = 1;
		for(int j = 0; j  maxsum){
				maxsum = cursum;
				maxid = i;
			} 
		}
		// 答案输出 
		printf("%d = ",n);
		for(int i = 0; i