Codeforces Round #395 A
题意:给你n,m求不超过z的即是n的倍数又是m的倍数的树有多少个
思路:z/(n*m) *gcd(n,m)小学奥数题,或者直接暴力也可以
AC代码:
#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; }