【codeforces 764A】Taymyr is calling you
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Comrade Dujikov is busy choosing artists for Timofey’s birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, …, z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
input
1 1 10
output
10
input
1 2 5
output
2
input
2 3 9
output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
【题目链接】:http://codeforces.com/contest/764/problem/A
【题解】
定义一个bool型的数组;
在n,2n,3n…设置true
然后在m,2m,3m处看看有没有为true的bool,有的话就递增答案;
(杀人真的很暴力。)
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int MAXN = 1e4+100;
int n,m,z,cnt=0;
bool bo[MAXN];
int main()
{
//freopen("F:\\rush.txt","r",stdin);
rei(n);rei(m);rei(z);
int i;
for (i = n;i<=z;i+=n)
bo[i] = true;
for (i = m;i<=z;i+=m)
if (bo[i])
cnt++;
printf("%d\n",cnt);
return 0;
}
【codeforces 764A】Taymyr is calling you的更多相关文章
- 【codeforces 415D】Mashmokh and ACM(普通dp)
[codeforces 415D]Mashmokh and ACM 题意:美丽数列定义:对于数列中的每一个i都满足:arr[i+1]%arr[i]==0 输入n,k(1<=n,k<=200 ...
- 【codeforces 750F】New Year and Finding Roots
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【codeforces 707E】Garlands
[题目链接]:http://codeforces.com/contest/707/problem/E [题意] 给你一个n*m的方阵; 里面有k个联通块; 这k个联通块,每个连通块里面都是灯; 给你q ...
- 【codeforces 707C】Pythagorean Triples
[题目链接]:http://codeforces.com/contest/707/problem/C [题意] 给你一个数字n; 问你这个数字是不是某个三角形的一条边; 如果是让你输出另外两条边的大小 ...
- 【codeforces 709D】Recover the String
[题目链接]:http://codeforces.com/problemset/problem/709/D [题意] 给你一个序列; 给出01子列和10子列和00子列以及11子列的个数; 然后让你输出 ...
- 【codeforces 709B】Checkpoints
[题目链接]:http://codeforces.com/contest/709/problem/B [题意] 让你从起点开始走过n-1个点(至少n-1个) 问你最少走多远; [题解] 肯定不多走啊; ...
- 【codeforces 709C】Letters Cyclic Shift
[题目链接]:http://codeforces.com/contest/709/problem/C [题意] 让你改变一个字符串的子集(连续的一段); ->这一段的每个字符的字母都变成之前的一 ...
- 【Codeforces 429D】 Tricky Function
[题目链接] http://codeforces.com/problemset/problem/429/D [算法] 令Si = A1 + A2 + ... + Ai(A的前缀和) 则g(i,j) = ...
- 【Codeforces 670C】 Cinema
[题目链接] http://codeforces.com/contest/670/problem/C [算法] 离散化 [代码] #include<bits/stdc++.h> using ...
随机推荐
- 2018-11-21-WPF-解决-ViewBox--不显示线的问题
title author date CreateTime categories WPF 解决 ViewBox 不显示线的问题 lindexi 2018-11-21 09:37:53 +0800 201 ...
- 三.使用JDBC处理MySql大数据
一.基本概念 大数据也称之为LOB(Large Objects),LOB又分为:clob和blob,clob用于存储大文本,blob用于存储二进制数据,例如图像.声音.二进制文等. 在实际开发中,有时 ...
- hdu5289 RMQ+二分
RMQ预处理最大值,最小值,然后对于每一点,二分可能满足的区间长度,长度-1就是该店开始的区间满足的个数. #include<stdio.h> #include<string.h&g ...
- 64位linux源码安装mysql
一:下载mysql http://dev.mysql.com/downloads/mysql/中的Generally Available(GA) Releases标签页,在MySQL Communit ...
- GIAC2019 演讲精选 | 面向未来的黑科技——UI2CODE闲鱼基于图片生成跨端代码
一直以来, 如何从‘视觉稿’精确的还原出 对应的UI侧代码 一直是端侧开发同学工作里消耗比较大的部分,一方面这部分的工作 比较确定缺少技术深度,另一方面视觉设计师也需要投入大量的走查时间,有大量无谓的 ...
- mysql字段中提取汉字,去除数字以及字母
如果只是删除尾部的中文,保留数据,可以用以下的简单方式 MySQL as num; +------+ | num | +------+ | +------+ DELIMITER $$ DROP FUN ...
- python 数据的读取
- mysql的root密码忘记解决方
mysql的root密码忘记解决方 没关注第一步,直接从第二步开始,(可以参看上一篇,先停止mysql服务).然后从第二步开始. 我启动mysql用的命令是/etc/init.d/mysql sta ...
- Spring AOP 的@Aspect
Spring AOP 的@Aspect 转自:http://blog.csdn.net/tanghw/article/details/3862987 从Spring 2.0开始,可以使用基于sch ...
- oracle函数 end
[功能]当:<表达式>=<表达式条件值1……n> 时,返回对应 <满足条件时返回值1……n> 当<表达式条件值1……n>不为条件表达式时,与函数deco ...