1103 Integer Factorization

题目：

The

Input Specification:

Each input file contains one test case which gives in a line the three positive integers

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

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

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<math.h>
using namespace std;
int n, k, p, maxFacSum = -1;
vector<int> ans, tempAns, v;
void init(){
    int tmp = 0, index = 1;
    while(tmp <= n){
        v.push_back(tmp);
        tmp = pow(index, p);
        index++;
    }
}
void dfs(int index, int tempSum, int tempK, int facSum){
    if(tempK == k){
        if(tempSum == n && facSum > maxFacSum){
            ans = tempAns;
            maxFacSum = facSum;
        }
        return;
    }
    while(index >= 1){
        if(tempSum + v[index] <= n){
            tempAns[tempK] = index;
            dfs(index, tempSum + v[index], tempK + 1, facSum + index);
        }
        if(index == 1) return;
        index--;
    }
}
 
int main(){
    cin>>n>>k>>p;
    init();
    tempAns.resize(k);
    dfs(v.size() - 1, 0, 0, 0);
    if(maxFacSum == -1){
        printf("Impossible");
    }else{
        cout<<n<<" = ";
        for(int i = 0; i < ans.size(); i++){
            cout<<ans[i]<<"^"<<p;
            if(i != ans.size() - 1){
                cout<<" + ";
            }
        }
    }
    return 0;
}

注意：

1、

while(index >= 1){
        if(tempSum + v[index] <= n){
            tempAns[tempK] = index;
            dfs(index, tempSum + v[index], tempK + 1, facSum + index);
        }
        if(index == 1) return;
        index--;
    }