PAT A1103 Integer Factorization （30 分）—

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 = 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 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

 #include <stdio.h>

 #include <algorithm>

 #include <set>

 #include <string.h>

 #include <vector>

 #include <math.h>

 #include <queue>

 using namespace std;

 bool cmp(int a,int b){

     return a>b;

 }

 const int maxn = ;

 int n,p,k,maxk=-;

 int vis[maxn]={};

 vector<int> res,tmp;

 void dfs(int index,int ksum,int cntk,int nsum){

     //if(index<1 || nsum>n || cntk>k) return;

     if(nsum==n && cntk == k){

         if(ksum>maxk){

             res=tmp;

             maxk=ksum;

         }

         return;

     }

     tmp.push_back(index);

     if(nsum+vis[index]<=n && cntk+<=k)dfs(index,ksum+index,cntk+,nsum+vis[index]);

     tmp.pop_back();

     if(index->)dfs(index-,ksum,cntk,nsum);

 }

 int main(){

     scanf("%d %d %d",&n,&k,&p);

     int i;

     for(i=;i<=n;i++){

         int res = pow(i,p);

         if(res>n)break;

         vis[i]=res;

     }

     i--;

     dfs(i,,,);

     if(maxk==-)printf("Impossible");

     else{

         printf("%d = ",n);

         sort(res.begin(),res.end(),cmp);

         for(int j=;j<res.size();j++){

             printf("%d^%d",res[j],p);

             if(j<res.size()-)printf(" + ");

         }

     }

 }

注意点：看到题目想到了要从大到小一个个遍历然后去比较条件，想用while和for写出来，发现真的写不来，有好多情况，看了大佬的思路，原来这就是递归，很明显的有个递归边界，递归式也很方便，果然对递归的理解还是不够深，知道思路，却没想到用递归这个武器。

巴特西

PAT A1103 Integer Factorization （30 分）——dfs，递归