【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
",cnt);
    return 0;
}