Codeforces Beta Round #67 (Div. 2)C. Modified GCD
2 seconds
256 megabytes
standard input
standard output
Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.
A common divisor for two positive numbers is a number which both numbers are divisible by.
But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor d between two integers a and b that is in a given range from low to high (inclusive), i.e. low ≤ d ≤ high. It is possible that there is no common divisor in the given range.
You will be given the two integers a and b, then n queries. Each query is a range from low to high and you have to answer each query.
The first line contains two integers a and b, the two integers as described above (1 ≤ a, b ≤ 109). The second line contains one integer n, the number of queries (1 ≤ n ≤ 104). Then n lines follow, each line contains one query consisting of two integers, low and high(1 ≤ low ≤ high ≤ 109).
Print n lines. The i-th of them should contain the result of the i-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.
9 27
3
1 5
10 11
9 11
3
-1
9
题目大意:找出a b的公因子 然后给几个查询,查询在x y之间的最大公因子 输出。
先把因子找出来,然后二分找一下最大的.
其实找因子 一共有两种方法(就我所知) 对于10^18这种级别的 用质因子那种方法找, 对于10^9这种级别的 用sqrt(n)的方法枚举因子的方法去找.....
/* ***********************************************
Author :guanjun
Created Time :2016/10/30 10:11:46
File Name :cf67c.cpp
************************************************ */
#include <bits/stdc++.h>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
int x,y;
};
struct cmp{
bool operator()(Node a,Node b){
if(a.x==b.x) return a.y> b.y;
return a.x>b.x;
}
}; bool cmp(int a,int b){
return a>b;
}
vector<int>v;
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
int a,b,x,y,n;
cin>>a>>b;
a=__gcd(a,b);
b=sqrt(a);
v.clear();
for(int i=;i<=b;i++){
if(a%i==){
if(i*i==a)v.push_back(i);
else v.push_back(i),v.push_back(a/i);
}
}
sort(v.begin(),v.end()); cin>>n;
for(int i=;i<=n;i++){
cin>>x>>y;
int pos=upper_bound(v.begin(),v.end(),y)-v.begin()-;
if(x>v[pos])puts("-1");
else cout<<v[pos]<<endl;
}
return ;
}
Codeforces Beta Round #67 (Div. 2)C. Modified GCD的更多相关文章
- Codeforces Beta Round #67 (Div. 2)
Codeforces Beta Round #67 (Div. 2) http://codeforces.com/contest/75 A #include<bits/stdc++.h> ...
- 【计算几何】 Codeforces Beta Round #67 (Div. 2) E. Ship's Shortest Path
读懂题意其实是模板题.就是细节略多. #include<cstdio> #include<cmath> #include<algorithm> using name ...
- Codeforces Beta Round #80 (Div. 2 Only)【ABCD】
Codeforces Beta Round #80 (Div. 2 Only) A Blackjack1 题意 一共52张扑克,A代表1或者11,2-10表示自己的数字,其他都表示10 现在你已经有一 ...
- Codeforces Beta Round #83 (Div. 1 Only)题解【ABCD】
Codeforces Beta Round #83 (Div. 1 Only) A. Dorm Water Supply 题意 给你一个n点m边的图,保证每个点的入度和出度最多为1 如果这个点入度为0 ...
- Codeforces Beta Round #79 (Div. 2 Only)
Codeforces Beta Round #79 (Div. 2 Only) http://codeforces.com/contest/102 A #include<bits/stdc++. ...
- Codeforces Beta Round #77 (Div. 2 Only)
Codeforces Beta Round #77 (Div. 2 Only) http://codeforces.com/contest/96 A #include<bits/stdc++.h ...
- Codeforces Beta Round #76 (Div. 2 Only)
Codeforces Beta Round #76 (Div. 2 Only) http://codeforces.com/contest/94 A #include<bits/stdc++.h ...
- Codeforces Beta Round #75 (Div. 2 Only)
Codeforces Beta Round #75 (Div. 2 Only) http://codeforces.com/contest/92 A #include<iostream> ...
- Codeforces Beta Round #74 (Div. 2 Only)
Codeforces Beta Round #74 (Div. 2 Only) http://codeforces.com/contest/90 A #include<iostream> ...
随机推荐
- MFC_2.2 编辑框和文本控件
编辑框和文本控件 1.拖控件 2.绑定变量.用户名密码编辑框控件类型.取名字.用户协议用值类型,默认CString. 设置属性.用户类型.选择mustiline TRUE. AOTO HScroll ...
- java byte
项目中有段代码,一直让我疑惑不解,但我是个很会偷懒的人,只要拷贝来改改能用的代码,万万不会自己动手写,虽然一直有疑惑,也懒得搭理是怎么个原理. 直到今天,又要解析协议,又要动这个地方的代码,还是来盘他 ...
- 12Microsoft SQL Server 索引
Microsoft SQL Server 索引 8.1创建索引 CREATE INDEX idx_name ON table_name(列名) --创建非聚集索引 use student go cre ...
- docker搭建日志收集系统EFK
EFK Elasticsearch是一个数据搜索引擎和分布式NoSQL数据库的组合,提过日志的存储和搜索功能. Fluentd是一个消息采集,转化,转发工具,目的是提供中心化的日志服务. Kibana ...
- jQuery元素节点的插入
jquery插入节点的的方法,总的来说有8种,但是只要学会了其中的两个就能理解全部了, 这里我们学习append()和appendTo()两个方法: append()方法是向元素的内部追加内容: &l ...
- Game Rank(NCPC 2016 大模拟)
题目: The gaming company Sandstorm is developing an online two player game. You have been asked to imp ...
- Educational Codeforces Round 57 (Rated for Div. 2) 前三个题补题
感慨 最终就做出来一个题,第二题差一点公式想错了,又是一波掉分,不过我相信我一定能爬上去的 A Find Divisible(思维) 上来就T了,后来直接想到了题解的O(1)解法,直接输出左边界和左边 ...
- python_ 学习笔记(基础语法)
python的注释 使用(#)对单行注释 使用('''或者""")多行注释,下面的代码肯定了python的牛逼 print("python是世界上最好的语言吗? ...
- 一篇入门MongoDB
目录 1.MongoDB 基本介绍 2.MongoDB 基本概念 3.数据库操作 4.集合操作 5.文档操作 6.查询条件 7.索引 1.MongoDB 基本介绍 (1)安装 MongoDB 简单来说 ...
- Python supprocess模块
当我们需要调用系统的命令的时候,最先考虑的os模块.用os.system()和os.popen()来进行操作.但是这两个命令过于简单,不能完成一些复杂的操作,如给运行的命令提供输入或者读取命令的输出, ...