PTA 甲级1103

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 (≤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

题目大意 ：

给定一个数字N，K，P，求是否有K个数字的P次方之和等于N，若存在，输出等式，若不存在，输出Impossible

题目分析：

在输出等式的时候要求找等式的加数和最大且整个序列最大的存在，所以应该使用深度优先搜索从上往下依次进行，并判断其和是否是最大值，若是最大值，则更新答案序列，否则继续进行搜索，在总和大于N以及搜索的层数到达K的时候注意剪枝。使用vector来存放数值有助于防止爆栈…

当时自己没有写出来，参考了柳小姐姐的博客，小姐姐的博客写的真好，一万个赞【链接】

代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include<algorithm>
#include <cmath>
#include<queue>
#include<map>
#include<stack>
#include<sstream>
using namespace std;
const int maxn = 1e5+7;
vector<int>v,ansque,tempque;
int n,k,p,maxque = -1;
bool exist = false;

int pows(int x, int y)
{
    int ans = 1;
    while(y>0)
    {

        if(y%2)
        {
            ans *= x;
        }
        x *= x;
        y >>= 1;
    }
    return ans;
}

void dfs(int index, int K, int sum, int que)
{
    //cout <<"index: " << index <<" K "<< K << " sum "<< sum << " que: "<< que << endl;
    if(K == k)
    {
        if(maxque < que && sum == n)
        {
            maxque = que;
            ansque = tempque;
            exist = true;
        }
        return;
    }
    for(int i = index; i > 0; i --)
    {
        //cout << i << " v[i]: " << v[i] << " sum: " << sum << endl;
        if(sum + v[i] <= n)
        {
            //cout << "zx" << endl;
            tempque[K] = i;
            dfs(i, K+1, sum+v[i],que+i);
        }
        //cout << i << " " << sum << endl;
    }
}

int main()
{
    cin >> n >> k >> p;
    int index = 1,temp = 0;
    while(temp <= n)
    {
        v.push_back(temp);
        temp = pows(index,p);
        index ++;
        //cout <<v[index-2] <<endl;
    }
    tempque.resize(k);
    dfs(index-2,0,0,0);
    if(exist)
    {
        cout << n << " = " << ansque[0] << "^" << p;
        for(int i = 1; i < k; i ++)
        {
            cout << " + " << ansque[i] << "^" << p;
        }
        cout << endl;

    }
    else
        cout << "Impossible" <<endl;
    return 0;
}