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 (≤), K (≤) and P (1). The numbers in a line are separated by a space.
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 1, or 1, 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 { , } is said to be larger than { , } if there exists 1 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
1 #include<iostream>
2 #include<cmath>
3 #include<vector>
4 using namespace std;
5 vector<int> temp, path;
6 int n, p, k, maxx=-1, t;
7 bool flag=true;
8 /*
9 dfs的回溯中弹出和压栈的位置很重要
10 */
11 void dfs(int a, int b, int c){
12 if(b==0 && c==0){
13 int tempMax=0;
14 for(int i=0; i<temp.size(); i++) tempMax+=temp[i];
15 if(maxx<tempMax){
16 maxx = tempMax;
17 path = temp;
18 }else if(maxx==tempMax && path<temp) path=temp;
19 temp.pop_back();
20 return;
21 }else if(c<=0 || b<=0){
22 temp.pop_back();
23 return;
24 }
25 for(int i=a; i>=1; i--){
26 if(b-c*pow(i, p)>0) break; //解决超时的关键点
27 if(b-pow(i, p)<0) {
28 while(b-pow(i, p)<0 && i>=1) --i;
29 temp.push_back(i);
30 dfs(i, b-pow(i, p), c-1);
31 }else{
32 temp.push_back(i);
33 dfs(i, b-pow(i, p), c-1);
34 }
35 }
36 temp.pop_back();//弹出的位置在for循环之外，很重要；
37 }
38 int main(){
39 scanf("%d%d%d", &n, &k, &p);
40 t =pow(n, 1.0/p) > n/k+1 ? pow(n, 1.0/p) : (n/k+1);
41 dfs(t, n, k);
42 if(path.size()==0) cout<<"Impossible"<<endl;
43 else {
44 printf("%d = ", n);
45 for(int i=0; i<path.size(); i++) {
46 if(i!=0) printf(" + ");
47 printf("%d^%d", path[i], p);
48 }
49 }
50 return 0;
51 }
原文地址：https://www.cnblogs.com/mr-stn/p/9584952.html