## 题目

The KP 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 KP 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:

 ``````1 `````` ``````N = n^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≤LK 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:

 ``````1 `````` ``````169 5 2 ``````

### Sample Output 1:

 ``````1 `````` ``````169 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2 ``````

### Sample Input 2:

 ``````1 `````` ``````169 167 3 ``````

### Sample Output 2:

 ``````1 `````` ``````Impossible ``````

## 代码

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 `````` ``````#include #include #include #include using namespace std; vector factor; vector ans; vector tmp; int N,K,P,fsum; int pow(int t,int pow) { int a=1; for(int i=0;iN) return; if(nowK==K) { if(powSum==N&&factSum>fsum) { fsum=factSum; ans=tmp; } return; } if(index>=1) { //select factor[index] tmp.push_back(index); DFS(index,nowK+1,factSum+index,powSum+factor[index]); tmp.pop_back(); //not select factor[index] DFS(index-1,nowK,factSum,powSum); } } int main() { scanf("%d%d%d",&N,&K,&P); int i,t; for(i=0;(t=pow(i,P))<=N;i++) { factor.push_back(t); } i--; DFS(i,0,0,0); if(fsum>0) { printf("%d = %d^%d",N,ans,P); for(i=1;i

## 分析

1. 预处理先计算出不大于N的所有\$i^p\$用于后续搜索
2. cmath中的pow函数返回值为double，在转换成int的时候是直接截断，例如10^2=100，在某些编译器下会出现(int)pow(10,2) => 99 ，所以只好自己写一个pow
3. DFS参数：当前搜索index，搜索到第nowK个数，因子和factSum，\$i^p\$ 之和powSum
4. 刚进入DFS的时候nowK是0
5. DFS的各种条件挺烦人的，还有vector在codebloks里无法调试，多注意吧