CodeForces#275--DIV 2--B(BinarySearch)(!!)
1 second
256 megabytes
standard input
standard output
You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.
In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.
Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.
A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.
The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Print a single integer — the answer to the problem.
3 1 2 3
5
1 3 2 3
4
In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4.
二分在最大求最小(whojay童鞋说的,我还没有查查,学渣伤不起)
#include <iostream> using namespace std; typedef long long ll; ll cur1, cur2, x, y; int main()
{
cin>>cur1>>cur2>>x>>y;
ll l = ;
ll r = 2e15;
while (l != r)
{
ll m = (l + r) >> 1LL;
ll f = m - (m/x);
ll s = m - (m/y);
ll all = m - (m/x/y);
if (f < cur1 || s < cur2 || all < cur1 + cur2)
l = m + ;
else
r = m;
}
cout<<l;
return ;
}
CodeForces#275--DIV 2--B(BinarySearch)(!!)的更多相关文章
- Codeforces Round #275 (Div. 2) C - Diverse Permutation (构造)
题目链接:Codeforces Round #275 (Div. 2) C - Diverse Permutation 题意:一串排列1~n.求一个序列当中相邻两项差的绝对值的个数(指绝对值不同的个数 ...
- Codeforces Round #275 (Div. 1)A. Diverse Permutation 构造
Codeforces Round #275 (Div. 1)A. Diverse Permutation Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 ht ...
- 构造 Codeforces Round #275 (Div. 2) C. Diverse Permutation
题目传送门 /* 构造:首先先选好k个不同的值,从1到k,按要求把数字放好,其余的随便放.因为是绝对差值,从n开始一下一上, 这样保证不会超出边界并且以防其余的数相邻绝对值差>k */ /*** ...
- Codeforces #344 Div.2
Codeforces #344 Div.2 Interview 题目描述:求两个序列的子序列或操作的和的最大值 solution 签到题 时间复杂度:\(O(n^2)\) Print Check 题目 ...
- Codeforces #345 Div.1
Codeforces #345 Div.1 打CF有助于提高做题的正确率. Watchmen 题目描述:求欧拉距离等于曼哈顿距离的点对个数. solution 签到题,其实就是求有多少对点在同一行或同 ...
- Codeforces Beta Round #27 (Codeforces format, Div. 2)
Codeforces Beta Round #27 (Codeforces format, Div. 2) http://codeforces.com/contest/27 A #include< ...
- Codeforces#441 Div.2 四小题
Codeforces#441 Div.2 四小题 链接 A. Trip For Meal 小熊维尼喜欢吃蜂蜜.他每天要在朋友家享用N次蜂蜜 , 朋友A到B家的距离是 a ,A到C家的距离是b ,B到C ...
- codeforces #592(Div.2)
codeforces #592(Div.2) A Pens and Pencils Tomorrow is a difficult day for Polycarp: he has to attend ...
- codeforces #578(Div.2)
codeforces #578(Div.2) A. Hotelier Amugae has a hotel consisting of 1010 rooms. The rooms are number ...
- codeforces #577(Div.2)
codeforces #577(Div.2) A Important Exam A class of students wrote a multiple-choice test. There are ...
随机推荐
- jquery获得select option的值 和对select option的操作
jQuery获取Select元素,并选择的Text和Value: 1. $("#select_id").change(function(){//code...}); //为Se ...
- printf 与c的基本类型之间的关系
型 書式 注意事項 ssize_t %zd size_t %zu intmax_t %jd uintmax_t %ju ptrdiff_t %t signed char %hhd unsigned c ...
- js获取url方法
//设置或获取对象指定的文件名或路径.alert(window.location.pathname); //设置或获取整个 URL 为字符串.alert(window.location.href); ...
- ViewPager部分源码分析一:加载数据
onMeasure()调用populate(),完成首次数据初始化. populate()维护ViewPager的page,包括mItems和mAdapter. populate(): if (cur ...
- 昨晚值班将发dla的程序改好后放入正式环境
可是在修改的topic的发送文件中出现有节点没有对应,整个过程陆续调至有20分钟最后11电把新程序换掉.
- JQuery阻止冒泡事件on绑定中异常情况分析
科普下事件冒泡以及默认行为,以下面例子举列子: 事件冒泡:当点击内部button元素时,会触发自身及外层 a的点击事件,这就是事件冒泡引起的.事件会随着 DOM 的层次结构依次向上传播. 事件 ...
- phpexcel中文教程-设置表格字体颜色背景样式、数据格式、对齐方式、添加图片、批注、文字块、合并拆分单元格、单元格密码保护
转:http://www.cnblogs.com/huangcong/p/3687665.html 首先到phpexcel官网上下载最新的phpexcel类,下周解压缩一个classes文件夹,里面包 ...
- 重温WCF之流与文件传输(七)
WCF开启流模式,主要是设置一个叫TransferMode的属性,所以,你看看哪些Binding的派生类有这个属性就可以了. TransferMode其实是一个举枚,看看它的几个有效值: Buffer ...
- BAT批量处理 命令
第一章 批处理基础第一节 常用批处理内部命令简介 批处理定义:顾名思义,批处理文件是将一系列命令按一定的顺序集合为一个可执行的文本文件,其扩展名为BAT或者CMD.这些命令统称批处理命令.小知识:可以 ...
- [Surface] 在win8.1上使用QQ截图放大问题(解决办法)
在使用每次截图的时候整个都被放大了,很让人郁闷,截不到完整的图,本着遇到问题解决问题的想法,这事早解决早好. 开工: 1. 度娘上搜索"win8 qq截图 放大",找到很多资料 ...