"Game theory" poj2348 Euclid ' s game

Assume the current b>a.

First, B%a==0 win

Second, b<2*a, at present, we have no choice, if the next step is to win (finally to the situation one), the current will be defeated, on the contrary, the current victory.

Third, b>2*a, assuming X is the integer that makes the b-ax<a, consider subtracting a (x-1) from B, for example, minus 12 for (4,19).

At this point, the next state becomes the former said there is no choice of circumstances two, if the state is a must, the current state is the winning state.

If the status is a win state, the next step is the only certainty, so it is a must, so we can directly reach this state.

∴ situation Three is the winning state.

∴ first reached the situation one or three is to win.

#include <cstdio> #include <algorithm>using namespace Std;long long A,b;int main () {while (1)  {  scanf ("%lld%lld", &a,&b);  if (!a&&!b) break;  BOOL Flag=1;while (1)    {    if (a>b) swap (A, b);  if (b%a==0| | B> (a<<1))    {    puts (flag?) Stan wins ":" Ollie wins ");    break;    }  B-=a;  flag^=1;    }  } return 0;}

"Game theory" poj2348 Euclid ' s game