Codeforces 764A Taymyr is calling you（水题翻译）

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本文解析了一个简单的算法竞赛题目——Codeforces A. Taymyriscallingyou，介绍了如何通过计算最小公倍数来解决该问题，并提供了完整的AC代码。

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.

题意：

Taymyr 打电话，每 n 分钟，同时每 m 分钟有一个艺术家过来被 kill？？（其实不会翻译）。问最后Taymyr打电话同时艺术家被 KO 的个数有几个。

思路：

求出最小公倍数即可，水题一枚，寒假算有AC记录了。

AC CODE：

#include<stdio.h>
#include<cstring>
#include<cmath>
#include<algorithm>
#define BadBoy main()
#define ForMyLove return 0;
using namespace std;
const int MYDD = 1103;

int gcd(int x, int y){
	if(!y) return x;
	return gcd(y, x%y);
}

int lcm(int x, int y){
	return x*y / gcd(x, y);
}

int BadBoy {
	int n, m, z;
	scanf("%d %d %d", &n, &m, &z);
	int ans = z / lcm(n, m);
	printf("%d\n", ans);
	ForMyLove
}