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 an…

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 in…

PAT甲级1103. Integer Factorization 题意: 正整数N的K-P分解是将N写入K个正整数的P次幂的和.你应该写一个程序来找到任何正整数N,K和P的N的K-P分解. 输入规格: 每个输入文件包含一个测试用例,它给出一行三个正整数N(<= 400), K(<= N)和P(1 <P <= 7).一行中的数字用空格分隔. 输出规格: 对于每种情况,如果解决方案存在,输出格式如下: N = n1 ^ P + ... nK ^ P 其中ni(i = 1,... K)是…

本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90574720 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…

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…

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 in…

暴力搜索. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<algorithm> using namespace std; int n,k,p,top; ]; ; ],ans_num=,ans_len; void d…

题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805364711604224 题意: 给定一个数n,要求从1~n中找出k个数,使得这些数的p次方之和等于n 思路: 因为n为400,所以dfs加剪枝[本来还在想dp来着] 他要求输出的方案是数字之和最大的,如果之和相等要输出字典序较大的. 所以还需维护一个数字之和. 一个剪枝的方法是从大到小进行dfs,后面被选的数一定比前面被选的要小,这里不限制的话显然会出现重复.…

该题还不错~. 题意:给定N.K.P,使得可以分解成N = n1^P + … nk^P的形式,如果可以,输出sum(ni)最大的划分,如果sum一样,输出序列较大的那个.否则输出Impossible. dfs枚举,为了防止超时,这里要预先将从1开始的i^p的值存储在factor数组中,直到i^p>n.然后dfs深度优先搜索,相当于把问题一步步分解,即若第一个因子是n1,则接下来我们要判断N-n1^p.k-1是否可行.同时存储当前因子的总和sum,要取sum最大的:还有上一次相加的因子的索引las…

题意: 输入三个正整数N,K,P(N<=400,K<=N,2<=P<=7),降序输出由K个正整数的P次方和为N的等式,否则输出"Impossible". //找到最大可能的整数pos后从大到小爆搜,sample 1给的输出好像不是最大的序列.....…