1103 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 (≤), K (≤) and P (1). 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 1, or 1, 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 { , } is said to be larger than { , } if there exists 1 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

我真的无语了，卡在测试点2卡了好久，怎么也没发现哪里不对，从刚写完代码的沾沾自喜到怎么调试也不过测试点2后上网搜索原因怀疑人生。结果发现是自己的==写成了！=，导致排序算法出错，沃柑！
错误真的是在你怎么也想不到的地方。

#include<iostream>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
vector<vector<int>>solutions;//存放所有的可能solution
int get_total(vector<int>a){
    int rt=0;
    for(int i=0;i<a.size();i++)rt+=a[i];
    return rt;
}
bool cmp(vector<int>a,vector<int>b){
    int ta=get_total(a);
    int tb=get_total(b);
    if(ta!=tb){
        return ta>tb;
    }else{
        int i=0;
        while(a[i]==b[i])i++;//把等号写成不等号在这里测试点2卡了很久
        return a[i]>b[i];
    }
}
vector<int>te;//临时解决方法；
int N,k,p;
int get_int(int a,int p){
    int te=1;
    for(int i=0;i<p;i++)te*=a;
    return te;
}
void DFS(int s,int tetotal,int p,int sk,int ed){
    if(sk==k&&tetotal==s){
        solutions.push_back(te);
        return;
    }
    if(tetotal==s&&sk<k)//数目不够
    {
        return;
    }
    for(int i=ed;i>=1;i--){
        if(tetotal+get_int(i,p)<=s&&sk+1<=k){
            te.push_back(i);
            DFS(s,tetotal+get_int(i,p),p,sk+1,i);//这里是一个剪枝，保证得到的序列从大到小_(:з」∠)_
            te.pop_back();
        }
    }
}
int main(){
    cin>>N>>k>>p;
    DFS(N,0,p,0,sqrt(N)+1);//因子的最大值不能比sqrt(N)+1还要大，所以遍历从sqrt(N)+1遍历就行了
    if(solutions.size()==0){
        cout<<"Impossible";
    }
    else{
        sort(solutions.begin(),solutions.end(),cmp);
        cout<<N<<" =";
        for(int i=0;i<k;i++){
            if(i!=0)cout<<" +";
            cout<<" "<<solutions[0][i]<<"^"<<p;
        }
    }
    return 0;
}