2^n的第一位数字 soj 3848 mathprac

发表于 2019-9-14 17:59

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.

发表于 2019-9-14 17:59

Input

* Line 1: Two space-separated integers: A and B

发表于 2019-9-14 17:59

Output

* Line 1: A single integer E calculated as above

发表于 2019-9-14 18:00

Sample Input

1 6

Sample Output

6

发表于 2019-9-14 18:00

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。

发表于 2019-9-14 18:00

复制

#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;

}

发表于 2019-9-14 18:01

作者：Matrix海子
　　　　
出处：http://www.cnblogs.com/dolphin0520/
　　　　
本博客中未标明转载的**归作者Matrix海子和博客园共有，欢迎转载，但未经作者同意必须保留此段声明，且在**页面明显位置给出原文连接，否则保留追究法律责任的权利。

2^n的第一位数字 soj 3848 mathprac

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