Put 100 BT_Us and 80 BT_Vs together, one minute later, there are 20 BT_Us and 80 BT_Vs, one minute later again, there are 20 BT_Us and 60 BT_Vs, then 20 BT_Us and 40 BT_Vs, then 20 BT_Us and 20 BT_Vs, then these 20 BT_Us and 20 BT_Vs keep alive.
Puts 3 BT_Us and 5 BT_Vs together, one minute later, there are 3 BT_Us and 2 BT_Vs, one more minute later there are 1 BT_U and 2 BT_Vs, then 1 BT_U and 1 BT_V, and this 1 BT_U and 1 BT_V keep alive.
According to the results above, Doctor Wang has reached a conclusion that when putting x BT_Us and y BT_Vs together, if x=y then they keep alive. if x>y then y BT_Us would die in one minute. Doctor Wang has mad a program to determine how many BT_Us and BT_Vs survive when putting x BT_Us and y BT_Vs together.
// Research for BT_U and BT_V
//===========================
#include"iostream"
using namespace std;
//---------------------------
int main(){
for(int x,y; cin>>x>>y; ){
while(x!=y)
x>y ? x-=y : y-=x;
cout<<x<<endl;
}//==========================