poj2115一次模余定理

http://poj.org/problem?id=2115

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<bitset>#include<iomanip>using namespace std;typedef __int64 INT ;INT ex_gcd( INT a , INT  b , INT &x ,INT &y ){if ( b == 0 ){x = 1 ; y = 0 ;return a ;}else{INT tm = ex_gcd( b , a % b , x , y ) ;INT t = x ;x = y ;y = t - ( a / b ) * y ;return tm ; }}int main(){ INT i , n , a1 ,r1 , a2 ,r2  ,ans , a , b , c , d , k , x0 , y0 ;while( scanf( "%I64d%I64d%I64d%I64d" , &a , &b , &c , &k ) , ( a + b + c + k ) ){INT temp = c ;c = b - a ;a = temp ;b = ( INT )1 << k ;d = ex_gcd( a , b , x0 , y0 ) ;if( c % d != 0 ){printf( "FOREVER\n" ) ;}else{INT ans = x0 * c / d ;INT temp = b / d ;ans = ( ans % temp + temp ) % temp ;printf( "%I64d\n" , ans ) ;}} return 0 ;}