[hdu 5051]2014上海网络赛 Fraction 数学 Benford's law/打表找规律

Fraction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 239    Accepted Submission(s): 55

Problem Description

Given a number n, and a geometric progression a _i = b * q _i, i ≥ 0, what is the fraction of the elements of that progression with decimal notation that has the decimal notation of n as prefix ?

More formally, if c _i out of the first i elements of the progression start with n in decimal notation, you need to find the limit . It is guaranteed that the limit always exists.

For example, n = 7, b = 1, q = 2. About 5.799% of all powers of two start with 7. (the smallest one is 2 ⁴⁶ = 70368744177664)

 

Input

The first line of the input is T (1 ≤ T ≤ 100), which stands for the number of test cases you need to solve.

Each case contains three integers n,b and q. (1 ≤ n, b, q ≤ 1000)

 

Output

For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then output one floating number – the sought fraction. Round your answer to the 5th decimal place.

 

Sample Input

   
   
   
   

    
    
    
    2
7 1 2
1 1 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 0.05799
Case #2: 1.00000
   
   
   
   

 

Source

2014 ACM/ICPC Asia Regional Shanghai Online

 

题目大意 

给定n,b,q

设 a _i  = b * q ^i

求当m趋近于无穷时，a1~am之中前缀为n的概率

解题思路

因为保证一定收敛，试着打表找一下规律

后来发现n取定后 不论b,q为何值都会基本收敛到一个数值，模拟100000项误差也不是很大

最后发现关于取定的n ans=(lg((n+1)/n))

特判一下q=1,10,100,1000的情况（相当于只向后面添加0，或者保持b不变）

想到了代码就水了，比赛的时候卡了多次边界= =

#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
  int T,ca=0;
  scanf("%d",&T);   
  while (T--)
  {
    int n,b,q;
    scanf("%d%d%d",&n,&b,&q);
    double nn=n,bb=b,qq=q;/*
    以下为打表
    bb=log10(bb);

    printf("Case #%d: ",++ca);
    int ans=0;
    for (int i=1;i<=1000000;i++)
    {
      bb+=log10(q);
      if ((int) (pow(10.0,(bb-((int)(bb))))+(int)(log10(double (n)))==n)
        if (bb>bb-((int)(bb))+(int)(log10(n)))
        ans++;
    }
    printf("%.5f\n",((double)ans)/1000000);
    打表结束
    */
    double r;

    if (q==10||q==100||q==1000) 
              if (b==n||b/10==n||b/100==n||b/1000==n||b*10==n||b*100==n||b*1000==n) r=1;
              else r=0;
    else
      if (q==1) if (b==n||b/10==n||b/100==n||b/1000==n) r=1;
            else r=0;
     else r=log10((double)(n+1)/double(n)); 
    printf("%.5f\n",r);
  }
  return 0;
}