题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1876
高精度的话辗转相除明显不靠谱，那么就用更相减损法来求最大公约数即可，然后练手一下高精度即可。
代码（压了9位的高精，跑得蛮快的）：
include <cstdio>
include <algorithm>
include <cstring>
using namespace std ;
define rep( i , x ) for ( int i = 0 ; i ++ < x ; )
define REP( i , l , r ) for ( int i = l ; i <= r ; ++ i )
define DOWN( i , r , l ) for ( int i = r ; i >= l ; -- i )
const int maxn = 1310 , maxb = 9 ;
int p[ maxb + 1 ] ;
inline void INIT( ) {
p[ 0 ] = 1 ; rep( i , maxb ) p[ i ] = p[ i - 1 ] * 10 ;
}
char s[ maxn * maxb ] ;
struct bignum {
int a[ maxn ] , n ;
bignum( ) {
memset( a , 0 , sizeof( a ) ) ;
}
inline void read( ) {
scanf( "%s" , s + 1 ) ;
int len = strlen( s + 1 ) , val ;
if ( len == 1 && s[ 1 ] == '0' ) a[ n = 1 ] = 0 ; else n = 0 ;
for ( int i = len ; i > 0 ; i -= maxb ) {
val = 0 ;
if ( i >= maxb ) REP( j , ( i - maxb + 1 ) , i ) val = val * 10 + s[ j ] - '0'
; else rep( j , i ) val = val * 10 + s[ j ] - '0' ;
a[ ++ n ] = val ;
}
}
inline void write( ) {
printf( "%d" , a[ n ] ) ;
DOWN( i , ( n - 1 ) , 1 ) {
REP( j , 1 , ( maxb - 1 ) ) if ( a[ i ] <= p[ j ] ) printf( "0" ) ;
printf( "%d" , a[ i ] ) ;
}
printf( "\n" ) ;
}
inline int cmp( const bignum &x ) {
if ( n > x.n ) return 1 ; else if ( n < x.n ) return - 1 ;
DOWN( i , n , 1 ) {
if ( a[ i ] > x.a[ i ] ) return 1 ; else if ( a[ i ] < x.a[ i ] ) return - 1 ;
}
return 0 ;
}
inline void div2( ) {
DOWN( i , n , 1 ) {
a[ i - 1 ] += ( a[ i ] & 1 ) * p[ maxb ] ; a[ i ] >>= 1 ;
}
if ( n > 1 && ! a[ n ] ) -- n ;
}
inline void mul2( ) {
rep( i , n ) a[ i ] <<= 1 ;
rep( i , n ) if ( a[ i ] >= p[ maxb ] ) {
a[ i + 1 ] += a[ i ] / p[ maxb ] ; a[ i ] %= p[ maxb ] ;
}
n += ( a[ n + 1 ] > 0 ) ;
}
inline void dec( const bignum &x ) {
rep( i , n ) {
for ( ; a[ i ] < x.a[ i ] ; a[ i ] += p[ maxb ] , -- a[ i + 1 ] ) ;
a[ i ] -= x.a[ i ] ;
}
DOWN( i , n , 1 ) if ( a[ i ] ) {
n = i ; break ;
}
}
inline bool odd( ) {
return a[ 1 ] & 1 ;
}
inline bool check( ) {
return a[ 1 ] == 0 && n == 1 ;
}
} n1 , n2 , ans , temp ;
int cnt = 0 ;
int main( ) {
INIT( ) ; n1.read( ) ; n2.read( ) ;
if ( n1.check( ) ) {
n2.write( ) ; return 0 ;
}
if ( n2.check( ) ) {
n1.write( ) ; return 0 ;
}
for ( int rec = n1.cmp( n2 ) ; rec ; rec = n1.cmp( n2 ) ) {
if ( n1.odd( ) ) {
if ( n2.odd( ) ) {
if ( rec == 1 ) n1.dec( n2 ) ; else n2.dec( n1 ) ;
} else n2.div2( ) ;
} else {
if ( n2.odd( ) ) n1.div2( ) ; else ++ cnt , n1.div2( ) , n2.div2( ) ;
}
}
ans = n1 ;
while ( cnt -- ) ans.mul2( ) ;
ans.write( ) ;
return 0 ;
}
