Taymyr is calling you

題意：給你n,m求不超過z的即是n的倍數又是m的倍數的樹有多少個

思路：z/(n*m) *gcd(n,m)小學奧數題，或者直接暴力也可以

AC代碼：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define
 
 ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a) memset(a,0,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

int n,m,z,ans;
 
int main(){
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n>>m>>z;
    cout<<z*__gcd(n，m)/(n*m)<<endl;
    return 0;
}

Codeforces Round #395 A