Time Limit: 3000 MS Memory Limit: 65536 K
mathprac
One lovely afternoon, Bessie's friend Heidi was helping Bessie
review for her upcoming math exam.
Heidi presents two integers A (0 <= A <= 45) and B (1 <= B <= 9)
to Bessie who must respond with an integer E in the range 1..62.
E is the smallest integer in that range that is strictly greater
than A and also has B as the first digit of 2 raised to the E-th
power. If there is no answer, Bessie responds with 0.
Help Bessie correctly answer all of Heidi's questions by calculating
her responses.
By way of example, consider A=1 and B=6. Bessie might generate a table
like this:
E 2^E First digit of 2^E
2 4 4
3 8 8
4 16 1
5 32 3
6 64 6 <-- matches B
Thus, E=6 is the proper answer.
NOTE: The value of 2^44 does not fit in a normal 32-bit integer.
* Line 1: Two space-separated integers: A and B
* Line 1: A single integer E calculated as above
1 6
mathprac
Description
One lovely afternoon, Bessie's friend Heidi was helping Bessie
review for her upcoming math exam.
Heidi presents two integers A (0 <= A <= 45) and B (1 <= B <= 9)
to Bessie who must respond with an integer E in the range 1..62.
E is the smallest integer in that range that is strictly greater
than A and also has B as the first digit of 2 raised to the E-th
power. If there is no answer, Bessie responds with 0.
Help Bessie correctly answer all of Heidi's questions by calculating
her responses.
By way of example, consider A=1 and B=6. Bessie might generate a table
like this:
E 2^E First digit of 2^E
2 4 4
3 8 8
4 16 1
5 32 3
6 64 6 <-- matches B
Thus, E=6 is the proper answer.
NOTE: The value of 2^44 does not fit in a normal 32-bit integer.
Input
* Line 1: Two space-separated integers: A and B
Output
* Line 1: A single integer E calculated as above
Sample Input
1 6
Sample Output
6
题意:
就是求2^n的第一位数字;
由于2^n=t*10^k,那么也就是求t的第一位。
log10(2^n)=n*log10(2)
=log10(t*10^k)
=log10(t)+k k为整数
求出n*log10(2)减去k即可求得log10(t)，然后求出10^log10(t)，对其取整，即为所求。
#include
<
iostream
>
#include < math.h >
using namespace std;
int main( void )
{
int A,B;
while (scanf( " %d%d " , & A, & B) == 2 )
{
int i;
double d;
int digit;
int p;
for (i = A + 1 ;i <= 62 ;i ++ )
{
d = i * log10( 2.0 );
p = ( int )(d);
digit = ( int )pow( 10.0 ,d - p);
if (digit == B)
{
printf( " %d\n " ,i);
break ;
}
}
if (i == 63 )
printf( " 0\n " );
}
return 0 ;
}
#include < math.h >
using namespace std;
int main( void )
{
int A,B;
while (scanf( " %d%d " , & A, & B) == 2 )
{
int i;
double d;
int digit;
int p;
for (i = A + 1 ;i <= 62 ;i ++ )
{
d = i * log10( 2.0 );
p = ( int )(d);
digit = ( int )pow( 10.0 ,d - p);
if (digit == B)
{
printf( " %d\n " ,i);
break ;
}
}
if (i == 63 )
printf( " 0\n " );
}
return 0 ;
}
