1103. Integer Factorization (30)-PAT甲级真题（dfs深度优先）

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 + … nK^P

where
ni (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 i<L and aL>bL

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

分析：dfs深度优先搜索。先把i从0开始所有的i的p次方的值存储在v[i]中，直到v[i]
> n为止。然后深度优先搜索，记录当前正在相加的index（即v[i]的i的值），当前的总和tempSum，当前K的总个数tempK，以及因为题目中要求输出因子的和最大的那个，所以保存一个facSum为当前因子的和，让它和maxFacSum比较，如果比maxFacSum大就更新maxFacSum和要求的ans数组的值。

在ans数组里面存储因子的序列，tempAns为当前深度优先遍历而来的序列，从v[i]的最后一个index开始一直到index
== 1，因为这样才能保证ans和tempAns数组里面保存的是从大到小的因子的顺序。一开始maxFacSum == -1，如果dfs后maxFacSum并没有被更新，还是-1，那么就输出Impossible，否则输出答案。

#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;
int n, k, p, maxFacSum = -1;
vector<int> v, ans, tempAns;
void init() {
int temp = 0, index = 1;
while(temp <= n) {
v.push_back(temp);
temp = pow(index, p);
index++;
}
}
void dfs(int index, int tempSum, int tempK, int facSum) {
if(tempSum == n && tempK == k) {
if(facSum > maxFacSum) {
ans = tempAns;
maxFacSum = facSum;
}
return ;
}
if(tempSum > n || tempK > k) return ;
if(index >= 1) {
tempAns.push_back(index);
dfs(index, tempSum + v[index], tempK + 1, facSum + index);
tempAns.pop_back();
dfs(index - 1, tempSum, tempK, facSum);
}
}
int main() {
scanf("%d%d%d", &n, &k, &p);
init();
dfs(v.size() - 1, 0, 0, 0);
if(maxFacSum == -1) {
printf("Impossible");
return 0;
}
printf("%d = ", n);
for(int i = 0; i < ans.size(); i++) {
if(i != 0) printf(" + ");
printf("%d^%d", ans[i], p);
}
return 0;
}