PAT 1103 Integer Factorization

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1103 Integer Factorization

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

总结：又是一道一点都不会的题目，真的想不到这个题目还能用DFS遍历的方法来写，长知识了，好好学习，多见多总结经验！

#include <iostream>
#include <cmath>
#include <vector>
using namespace std;

int n,k,p;
int maxfacSum=-1;
vector<int> v,ans,temp;
void init(){
    int temp=0,index=1;
    while(temp<=n){
        v.push_back(temp);
        temp=pow(index,p);
        index++;
    }
}
//tempK表示使用了多少个数，facSum表示这些数的总和 
void dfs(int index,int temSum,int temk,int facSum){
    if(temk==k){
        if(temSum==n && facSum>maxfacSum){
            ans=temp;
            maxfacSum=facSum;
        }
        return;
    }
    while(index>=1){
        if(temSum+v[index]<=n){
            temp[temk]=index;
            //dfs index不将index--表示可以重复使用同一个数 
            dfs(index,temSum+v[index],temk+1,facSum+index);
        }
        if(index==1)    return;//为什么需要大于1，是因为v[0]存的是0，v[1]存的才是1
        index--;
    }
}
int main(){
    scanf("%d%d%d",&n,&k,&p);
    temp.resize(k);
    init();
    dfs(v.size()-1,0,0,0);
    if(maxfacSum==-1)   puts("Impossible");
    else{
        printf("%d = %d^%d",n,ans[0],p);
        for(int i=1;i<ans.size();i++)
            printf(" + %d^%d",ans[i],p);
    }
    return 0;
}

好好学习，天天向上！

我要考研！