PAT A1103 Integer Factorization】的更多相关文章

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…
线性dfs,注意每次深搜完状态的维护~ #include<bits/stdc++.h> using namespace std; ; vector<int> v,tmp,path; int n,k,p; void init () { ,cnt=; while (t<=n) { v.push_back(t); t=pow(cnt,p); cnt++; } } ; void dfs (int nowindex,int nowsum,int nowK,int facSum) { i…
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 Kpositive 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 inp…
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…
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…
Source: PAT A1103 Integer Factorization (30 分) Description: The K−P factorization of a positive integer N is to write N as the sum of the P-th power of Kpositive integers. You are supposed to write a program to find the K−P factorization of N for any…
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…
PAT A1103 标签(空格分隔): PAT 解题思路: DFS #include <cstdio> #include <vector> using namespace std; int n, k, p; int maxFacSum = -1; //记录最大因子和 vector<int> ans, temp, fac; int power(int x) { int ans = 1; for(int i = 0; i < p; i++) { ans *= x; }…
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…
 Integer Factorization 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…
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.....…
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…
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…
Given a set of N (>1) positive integers, you are supposed to partition them into two disjoint sets A​1​​ and A​2​​ of n​1​​ and n​2​​ numbers, respectively. Let S​1​​ and S​2​​ denote the sums of all the numbers in A​1​​ and A​2​​, respectively. You…
Given a set of N (>1) positive integers, you are supposed to partition them into two disjoint sets A1 and A2 of n1​​ and n​2 numbers, respectively. Let S1​ and S​2 denote the sums of all the numbers in A​1​​ and A​2​​ , respectively. You are supposed…
题意:给出一个正整数N,以及k,p,要求把N分解成k个因式,即N=n1^p + n2^p + ... + nk^p.要求n1,n2,...按降序排列,若有多个解,则选择n1+n2+...+nk最大的,若还有多个解,则选择数字序列较大的.若不存在解,则输出Impossible. 思路:这是深度优先搜索的经典例题.详见: 代码: #include <cstdio> #include <cstdlib> #include <cmath> #include <vector…
1091 Acute Stroke (30point(s)) 基础的搜索,但是直接用递归会导致段错误,改用队列之后就不会了,这说明递归调用在空间利用率上还是很吃亏的. #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <iostream> #include <cmath> #include <map> #…
2019/4/3 1063 Set Similarity n个序列分别先放进集合里去重.在询问的时候,遍历A集合中每个数,判断下该数在B集合中是否存在,统计存在个数(分子),分母就是两个集合大小减去分子. // 1063 Set Similarity #include <set> #include <map> #include <cstdio> #include <iostream> #include <algorithm> using name…
博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…
树(23) 备注 1004 Counting Leaves   1020 Tree Traversals   1043 Is It a Binary Search Tree 判断BST,BST的性质 1053 Path of Equal Weight   1064 Complete Binary Search Tree 完全二叉树的顺序存储,BST的性质 1066 Root of AVL Tree 构建AVL树,模板题,需理解记忆 1079 Total Sales of Supply Chain…
最短路径 Emergency (25)-PAT甲级真题(Dijkstra算法) Public Bike Management (30)-PAT甲级真题(Dijkstra + DFS) Travel Plan (30)-PAT甲级真题(Dijkstra + DFS,输出路径,边权) All Roads Lead to Rome (30)-PAT甲级真题-Dijkstra + DFS Online Map (30)-PAT甲级真题(Dijkstra + DFS) 最短路径扩展问题 要求数最短路径有多…
专题一  字符串处理 A1001 Format(20) #include<cstdio> int main () { ]; int a,b,sum; scanf ("%d %d",&a,&b); sum=a+b; ) { printf ("-"); sum=-sum; } ; ) { s[top++]=; } ) { s[top++]=sum%; sum/=; } ;i>=;i--) { printf ("%d"…
Source: PAT (Advanced Level) Practice Reference: [1]胡凡,曾磊.算法笔记[M].机械工业出版社.2016.7 Outline: 基础数据结构: 线性表:栈,队列,链表,顺序表 树:二叉树的建立,二叉树的遍历,完全二叉树,二叉查找树,平衡二叉树,堆,哈夫曼树 图:图的存储和遍历 经典高级算法: 深度优先搜索,广度优点搜索,回溯剪枝 贪心,并查集,哈希映射 最短路径(只考察过单源),拓扑排序(18年9月第一次涉及相关概念,未正式考过),关键路径(未…
Source: PAT A1113 Integer Set Partition (25 分) Description: Given a set of N (>) positive integers, you are supposed to partition them into two disjoint sets A​1​​and A​2​​ of n​1​​ and n​2​​ numbers, respectively. Let S​1​​ and S​2​​ denote the sums…