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

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 12​2​​+4​2​​+2​2​​+2​2​​+1​2​​, or 11​2​​+6​2​​+2​2​​+2​2​​+2​2​​, 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 { a​1​​,a​2​​,⋯,a​K​​ } is said to be larger than { b​1​​,b​2​​,⋯,b​K​​ } if there exists 1≤L≤K 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

 #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。