Time Limit: 3000 MS Memory Limit: 65536 K
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 ; } http://www.cnblogs.com/dolphin0520/archive/2011/04/11/2012982.html