Educational Codeforces Round 13 C. Joty and Chocolate 水题
C. Joty and Chocolate
题目连接:
http://www.codeforces.com/contest/678/problem/C
Description
Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.
An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.
After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.
Note that she can paint tiles in any order she wants.
Given the required information, find the maximum number of chocolates Joty can get.
Input
The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).
Output
Print the only integer s — the maximum number of chocolates Joty can get.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Sample Input
5 2 3 12 15
Sample Output
39
Hint
题意
有1到n n个数,如果k%a0,那么给p元,如果k%b0,那么给q元
如果k%a0||k%b0,那么给p元或者q元
问你最多给多少
题解:
k%a0||k%b0 > k%lcm(a,b)0 给max(p,q)元
然后搞一波就好了……
代码
#include<bits/stdc++.h>
using namespace std;
long long gcd(long long a,long long b)
{
if(b==0)return a;
return gcd(b,a%b);
}
long long lcm(long long a,long long b)
{
return a*b/gcd(a,b);
}
long long n,a,b,p,q;
int main()
{
cin>>n>>a>>b>>p>>q;
long long ans1 = n/a;
long long ans2 = n/b;
long long ans3 = n/lcm(a,b);
ans1-=ans3,ans2-=ans3;
cout<<ans1*p+ans2*q+(ans3)*max(p,q)<<endl;
}
Educational Codeforces Round 13 C. Joty and Chocolate 水题的更多相关文章
- Educational Codeforces Round 13 D. Iterated Linear Function 水题
D. Iterated Linear Function 题目连接: http://www.codeforces.com/contest/678/problem/D Description Consid ...
- Educational Codeforces Round 13 B. The Same Calendar 水题
B. The Same Calendar 题目连接: http://www.codeforces.com/contest/678/problem/B Description The girl Tayl ...
- Educational Codeforces Round 13 A. Johny Likes Numbers 水题
A. Johny Likes Numbers 题目连接: http://www.codeforces.com/contest/678/problem/A Description Johny likes ...
- Educational Codeforces Round 14 A. Fashion in Berland 水题
A. Fashion in Berland 题目连接: http://www.codeforces.com/contest/691/problem/A Description According to ...
- Codeforces Round #340 (Div. 2) B. Chocolate 水题
B. Chocolate 题目连接: http://www.codeforces.com/contest/617/problem/D Descriptionww.co Bob loves everyt ...
- Educational Codeforces Round 4 A. The Text Splitting 水题
A. The Text Splitting 题目连接: http://www.codeforces.com/contest/612/problem/A Description You are give ...
- Codeforces Educational Codeforces Round 3 B. The Best Gift 水题
B. The Best Gift 题目连接: http://www.codeforces.com/contest/609/problem/B Description Emily's birthday ...
- Codeforces Educational Codeforces Round 3 A. USB Flash Drives 水题
A. USB Flash Drives 题目连接: http://www.codeforces.com/contest/609/problem/A Description Sean is trying ...
- Educational Codeforces Round 12 A. Buses Between Cities 水题
A. Buses Between Cities 题目连接: http://www.codeforces.com/contest/665/problem/A Description Buses run ...
随机推荐
- PostgreSQL内核分析——BTree索引
文中附图参考至<PostgreSQL数据库内核分析> (一)概念描述 B+树是一种索引数据结构,其一个特征在于非叶子节点用于描述索引,而叶子节点指向具体的数据存储位置.在PostgreSQ ...
- Ubuntu下安装arm-linux-gnueabi-xxx编译器【转】
转自:http://blog.csdn.net/real_myth/article/details/51481639 from: http://www.linuxdiyf.com/linux/1948 ...
- gcc -O0 -O1 -O2 -O3 四级优化选项及每级分别做什么优化【转】
转自:http://blog.csdn.net/qinrenzhi/article/details/78334677 相关博客http://blog.chinaunix.net/uid-2495495 ...
- aarch64_o2
opensips-event_rabbitmq-2.2.3-1.fc26.aarch64.rpm 2017-03-10 01:22 42K fedora Mirroring Project opens ...
- Awk基础
Awk文本处理 awk是一种编程语言,用于在linux/unix下对文本和数据进行处理.awk数据可以来自标准输入.一个或多个文件,或其它命令的输出.awk通常是配合脚本进行使用, 是一个强大的文本处 ...
- orcale数据库分配用户
account lock:创建用户的时候锁定用户 account unlock:创建用户的时候解锁用户,默认该选项 create user zhou8–用户名 identified by zhou88 ...
- c语言双向循环链表
双向循环链表,先来说说双向链表,双向链表也叫双链表,是链表的一种,它的每个数据结点中都有两个指针,分别指向直接后继和直接前驱.所以,从双向链表中的任意一个结点开始,都可以很方便地访问它的前驱结点和后继 ...
- Winafl学习笔记
最近在跟师傅们学习Winafl,也去搜集了一些资料,有了一些自己的理解,就此记录一下. Winafl是一个运行时插桩工具,可以提高crash的捕获率. 同时也有自己的遗传算法,可以根据代码覆盖程度进行 ...
- Django实战(20):分页(Pagination)
在上一节我们实现了针对某个产品的订单订阅功能.但是我们可能需要直接在站点上查询所有的订单.显然,随着时间的增长订单会越来越多,所以分页(Pagination)是个好办法:每次只显示一部分订单. 分页是 ...
- TCP、UDP、HTTP、SOCKET之间的区别与联系
IP:网络层协议: TCP和UDP:传输层协议: HTTP:应用层协议: SOCKET:TCP/IP网络的API. TCP/IP代表传输控制协议/网际协议,指的是一系列协议. TCP和UDP使用IP协 ...