2^n的第一位数字 soj 3848 mathprac

    Time Limit: 3000 MS    Memory Limit: 65536 K

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                                                      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;
}