HDOJ 3641 Treasure Hunting

HDOJ 3641 Treasure Hunting

给出 2n 个数，求最大的 x 满足 x!%M = 0 ，其中 M = a1^b1*a2^b2*a3^b3…*an^bn 。

Input

In the first line is an integer T (1<=T<=50) indicating the number of test cases.
Each test case begins with an integer n (1<=n<=100), then followed n lines. Each line contains two numbers ai and bi (1 <= ai <= 100, 1<=bi<=10000000000000)

Output

For each test case output the result x in one line.

Source
2010 Asia Regional Hangzhou Site Online Contest

# include  < stdio.h >
# include  < string .h >
# include  < math.h >

# define N  101
typedef  long   long   int  LL;

int  n;
/* ***************************************************** */
int  p[ 50 ], top  =   0 ;
int  isPrime( int  x)
{
                 int  i;
                 if  (x % 2 == 0 )  return  x == 2 ;
                 if  (x % 3 == 0 )  return  x == 3 ;
                 if  (x % 5 == 0 )  return  x == 5 ;
                 for  (i  =   7 ; i  <  x; i  +=   5 )
                                 if  (x % i  ==   0 )  return   0 ;
                 return   1 ;
}

void  pre( void )
{
                 int  i;
                 for  (i  =   2 ; i  <  N;  ++ i)
                                 if  (isPrime(i)) p[top ++ ]  =  i;
}
/* ***************************************************** */
LL pn[ 50 ];
void  init( void )
{
                 int  i, j, x;
                LL cnt;
                LL num;
                memset(pn,  0 ,  sizeof (pn));
                scanf( " %d " ,  & n);
                 for  (i  =   0 ; i  <  n;  ++ i)
                {
                                scanf( " %d%I64d " ,  & x,  & num);
                                 for  (j  =   0 ; j  <  top;  ++ j)
                                {
                                                 if  (x % p[j]  ==   0 )
                                                {
                                                                cnt  =   0 ;
                                                                 while  (x % p[j] == 0 )  ++ cnt, x /= p[j];
                                                                pn[j]  +=  cnt * num;
                                                }
                                }
                }
}
LL Max(LL x, LL y)
{
                 return  x > y  ?  x:y;
}

LL mypow( int  pr,  int  cnt)
{
                LL ret  =   1 ;
                 while  (cnt  >   0 )  -- cnt, ret  *=  pr;
                 return  ret;
}

LL cal( int  pr, LL tot)
{
                 int  tmp;
                LL ppow  =   0 , temp;
                 while  (tot  >   0 )
                {
                                tmp  =  ( int )floor(log(tot * (pr - 1 ) + 1 ) / log(pr)) + 1 ;
                                 while  ((mypow(pr, tmp) - 1 ) / (pr - 1 )  >  tot)  -- tmp;
                                temp  =  mypow(pr, tmp);
                                ppow  +=  temp;
                                tot  -=  (temp - 1 ) / (pr - 1 );
                }
                 return  ppow;
}
void  solve( void )
{
                 int  i;
                LL ans  =   0 ;
                 for  (i  =   0 ; i  <  top;  ++ i)  if  (pn[i]  !=   0 )
                                                ans  =  Max(ans, cal(p[i], pn[i]));
                printf( " %I64d\n " , ans);
}

int  main()
{
                 int  T;
                pre();
                scanf( " %d " ,  & T);
                 while  (T -- ) init(), solve();

                 return   0 ;
}