PAT (Advanced Level) Practise 1103 Integer Factorization (30)-优快云博客

本文链接：https://blog.youkuaiyun.com/a664607530/article/details/73991966

本文介绍了一种通过深度优先搜索加剪枝策略解决特定整数K-P分解问题的方法。目标是在给定正整数N、K及P的情况下，寻找K个正整数，使它们的P次幂之和等于N，并在多种解中找到元素和最大且序列最大的解。

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1103. Integer Factorization (30)

时间限制

1200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 = n₁^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, ... a_K } is said to be larger than { b₁, b₂, ... 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

题意：给你一个n，k，p，让你找出一个有k个元素的序列，它们p次方的和等于n，若有多个，输出元素和最大的，若仍有多个，输出序列最大的

解题思路：dfs+剪枝

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x7FFFFFFF;

int n,k,p,ans[500],flag,a[500],ma,b[500];

int mypow(int x,int k)
{
    int ans=1;
    while(k)
    {
        if(k&1) ans*=x;
        k>>=1;
        x*=x;
    }
    return ans;
}

void dfs(int x,int sum,int pos)
{
    if(sum==n&&k==x)
    {
        int sum1=0;
        for(int i=0;i<k;i++) sum1+=a[i],b[i]=a[i];
        sort(b,b+k,greater<int>());
        int p=0;
        while(b[p]==ans[p]&&p<k)p++;
        if((b[p]>ans[p]&&p<k&&sum1==ma)||sum1>ma)
        {
            ma=sum1;
            for(int i=0;i<k;i++) ans[i]=b[i];
        }
        flag=1;
        return ;
    }
    int xx=mypow(pos,p);
    if(sum+(k-x)*xx>n||x>=k||sum>=n) return ;
    for(int i=pos; i<=20; i++)
    {
        a[x]=i;
        dfs(x+1,sum+mypow(i,p),i);
    }
}

int main()
{
    while(~scanf("%d%d%d",&n,&k,&p))
    {
        flag=0;
        ma=0;
        dfs(0,0,1);
        if(!flag) {printf("Impossible\n");continue;}
        printf("%d = %d^%d",n,ans[0],p);
        for(int i=1; i<k; i++)
            printf(" + %d^%d",ans[i],p);
        printf("\n");
    }
    return 0;
}