1103 Integer Factorization (30分)

题目地址

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<iostream>
#include<algorithm>
#include<vector>
#include<stdlib.h>
#include<math.h>
using namespace std;
vector<int> fac,ans,temp;
int n,k,p,max_fac_sum=-1;
void dfs(int index,int now_k,int sum,int fac_sum){
    if(sum==n&&now_k==k){           //因子数量到达要求，项和恰好为n到达递归终点；
        if(fac_sum>max_fac_sum){            //所有满足的因子，贪心要底数更大的；不判断少五分；
            ans=temp;max_fac_sum=fac_sum;
        }
        return;
    }
    if(sum>n||now_k>k||index<1) return;          //递归终点，数量超过要求，项和超过要求,某项底数变成0；
    temp.push_back(index);          //优先把最大的底数全部放入temp，然后一项一项拿出，将次等底数填入；
    dfs(index,now_k+1,sum+fac[index],fac_sum+index);       //不管判断结果如何，上面有递归终点和大小判断，只需枚举就行；
    temp.pop_back();
    dfs(index-1,now_k,sum,fac_sum);     //拿出一项，填入次等底数，总数量不变；
}
int main(){
    scanf("%d%d%d",&n,&k,&p);
    int i=0,t=0;
    while(t<=n){
        fac.push_back(t);t=pow(++i,p);        //pow ,no 
    }
    dfs(fac.size()-1,0,0,0);
    if(max_fac_sum==-1) printf("Impossible\n");
    else{
        printf("%d = %d^%d",n,ans[0],p);
        for(int i=1;i<ans.size();i++){
            printf(" + %d^%d",ans[i],p);
        }
    }
}

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