Codeforces Round #596 (Div. 2)D.Power Products
题意:
给一个数组,给你一个k,找出两个数字的积可以变成xk的数对对数
解析:
当且仅当,两个数进行质因子分解后每个因子的个数都是k的倍数个就说明这是满足条件的一对,可以让每个因子个数%k用map找对应的数。
代码:
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <cstdio>
#include <queue>
#include <cmath>
#include <map> using namespace std; typedef long long LL; const int mod=1e9+;
const int maxn=1e5+; int n,k; struct node{
int x,num;
node(){}
node(int _x,int _num):x(_x),num(_num){}
bool operator<(const node &s)const{
if(x!=s.x)return x<s.x;
else return num<s.num;
}
}; vector<node>nodes;
map<vector<node>,int>mp; void f(int x){
int o=;
for(int i=;i<=x/i;i++){
if(x%i==){
int num=;
while(x%i==){
num++;
x/=i;
}
if(num%k!=){
o=;
nodes.push_back(node(i,num%k));
}
}
}
if(x!=)nodes.push_back(node(x,));
if((int)nodes.size()==&&(o==||x==)){
//nodes.push_back(node(x,1));
}
} int main(){
scanf("%d%d",&n,&k);
long long num=;
for(int i=;i<n;i++){
int x;
scanf("%d",&x);
nodes.clear();
f(x);
for(int i=;i<(int)nodes.size();i++){
nodes[i].num=k-nodes[i].num;
}
if(mp[nodes]!=){
num+=mp[nodes];
}
for(int i=;i<(int)nodes.size();i++){
nodes[i].num=k-nodes[i].num;
//printf("%d %d\n",nodes[i].x,nodes[i].num);
}
mp[nodes]++;
//printf("%d %lld\n",i,num);
}
printf("%lld\n",num);
return ;
}
Codeforces Round #596 (Div. 2)D.Power Products的更多相关文章
- Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2)
A - Forgetting Things 题意:给 \(a,b\) 两个数字的开头数字(1~9),求使得等式 \(a=b-1\) 成立的一组 \(a,b\) ,无解输出-1. 题解:很显然只有 \( ...
- Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) D. Power Products 数学 暴力
D. Power Products You are given n positive integers a1,-,an, and an integer k≥2. Count the number of ...
- Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) D. Power Products
链接: https://codeforces.com/contest/1247/problem/D 题意: You are given n positive integers a1,-,an, and ...
- Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) C. p-binary
链接: https://codeforces.com/contest/1247/problem/C 题意: Vasya will fancy any number as long as it is a ...
- Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) C. p-binary 水题
C. p-binary Vasya will fancy any number as long as it is an integer power of two. Petya, on the othe ...
- Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) B2. TV Subscriptions (Hard Version)
链接: https://codeforces.com/contest/1247/problem/B2 题意: The only difference between easy and hard ver ...
- Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) A. Forgetting Things
链接: https://codeforces.com/contest/1247/problem/A 题意: Kolya is very absent-minded. Today his math te ...
- Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) F. Tree Factory 构造题
F. Tree Factory Bytelandian Tree Factory produces trees for all kinds of industrial applications. Yo ...
- Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) E. Rock Is Push dp
E. Rock Is Push You are at the top left cell (1,1) of an n×m labyrinth. Your goal is to get to the b ...
随机推荐
- Go1.14发布了,快来围观新的特性啦
如期而至,Go1.14发布了,和往常一样,该版本保留了Go 1兼容性的承若,这个版本的大部分更新在工具链 .运行时库的性能提升方面,总的来说,还是在已有的基础上不断优化提成,大家期待的泛型还没有到来, ...
- .net List回收
转 static void Main(string[] args) { List<int> list = new List<int>(); for (int i = 0; i ...
- UML之二、建模元素(1)
本章介绍UML建模元素 1:Stereotype-也被称为类型.构造型 UML里的元素扩展,简单来说其功能就是在已有的类型上添加一些标记,类似于打个戳,从而生成新的东西. 简单的说加一句话来更加清楚准 ...
- 链接github
引用https://www.cnblogs.com/u-1596086/p/11588957.html 第一步:登录git创建项目 右上角头像按钮,点击your repositories 接着绿色按钮 ...
- Flink中逻辑计划和物理计划的概念划分和对应关系
逻辑计划 logicGraph或者jobGraph,其端点为operator,edge为数据流向. operator往往代表一个函数. 同一个分区内的具有连续上下游关系的函数组成operator-ch ...
- 基于Jupyter Notebooks的C# .NET Interactive安装与使用
.NET Interactive发布预览版了,可以像Python那样用jupyter notebooks来编辑C#代码.具体可以在GitHub上查看dotnet/interactive项目. 安装步骤 ...
- python爬虫步骤 (新手备学 )爬虫编程。
Python爬虫是用Python编程语言实现的网络爬虫,主要用于网络数据的抓取和处理,相比于其他语言,Python是一门非常适合开发网络爬虫的编程语言,大量内置包,可以C Python爬虫可以做的事情 ...
- 使用TensorRT对人脸检测网络MTCNN进行加速
前言 最近在做人脸比对的工作,需要用到人脸关键点检测的算法,比较成熟和通用的一种算法是 MTCNN,可以同时进行人脸框选和关键点检测,对于每张脸输出 5 个关键点,可以用来进行人脸对齐. 问题 刚开始 ...
- jmeter-json提取器提取的内容含”引号
这时如果直接赋值会报错 解决方法: 1.用vars.get("Object")提取变量的值 2.用代码提取,最后把提取到的Object或Array转为String
- 基于JavaSwing开发银行信用卡管理系统
开发环境: Windows操作系统开发工具: MyEclipse10/Eclipse+Jdk+Mysql数据库 运行效果图 源码及原文链接:https://javadao.xyz/forum.php? ...