1103 Integer Factorization (30 分) dfs 剪枝

忘是亡心i 2021-11-02 00:48 288阅读 0赞

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

这道题题意很简单，给出s,n,k找一组数满足s=n个数的k次方之和

如果有多组数满足，则取底数之和最大的，且输出的时候底数从大到小输出。

深搜里记录：x-当前值，nown-当前共有n个数，nowsum-当前和

当nown==n&&nowsum==s时满足条件，比较sum和res\_sum，如果更大，则替换结果res=v;

#include <stdio.h>
    #include <string.h>
    #include <string>
    #include <map>
    #include <queue>
    #include <math.h>
    #include <algorithm>
    #include <iostream>
    #define INF 0x3f3f3f3f
    using namespace std;
    int s,n,k,res_sum;
    vector<int> v,res;
    void dfs(int x,int nown,int nowsum)
    {
    	if(x<=0||nown>n||nowsum>s)return;
    	//printf("%d %d %d\n",x,nown,nowsum);
    	if(nown==n&&nowsum==s)
    	{
    		int sum=0;
    		for(int i=0;i<v.size();i++)
    		sum+=v[i];
    		//printf("%d\n",sum);
    		if(sum>res_sum)
    		{
    			res_sum=sum;
    			res=v;
    		}
    		return;
    	}
    	v.push_back(x);
    	dfs(x,nown+1,nowsum+pow(x,k));
    	v.pop_back();
    	dfs(x-1,nown,nowsum);
    }
    int main()
    {
    	scanf("%d%d%d",&s,&n,&k);
    	res_sum=-1;
    	dfs(sqrt(s),0,0);
    	if(res_sum!=-1)
    	{
    		printf("%d = ",s);
    		sort(res.begin(),res.end());
    		for(int i=res.size()-1;i>0;i--)
    		printf("%d^%d + ",res[i],k);
    		printf("%d^%d\n",res[0],k);
    	}	
    	else
    	printf("Impossible\n");
    }