Logarithms (数学)

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本文探讨了数值计算中的对数函数，通过特定公式求解对数表达式的最小可能值及对应变量x的值，涉及数学公式应用与计算技巧。

Logarithms

Time Limit:1000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status Practice UVA 11666

Description

Problem A
Logarithms
Input: Standard Input

Output: Standard Output

From time immemorial different series has been an integrated part of mathematics. Series is very important for finding values of many important functions such as sin(x), e^x, ln(x) etc. The well known formula for finding the value of ln(1-x) is shown below:

, Here |x|<1.

However as this formula is true when x is less than 1, a modification is needed to find the formula for any integer. For any integer n the following relationship is true:

Here |x|<1 and it is a real number, n is a positive integer and L is a non-negative integer.

But for a given integer n, L can have more than one value. Your job is to find the smallest possible value of L and for that L find the value of x.

Input

The input file contains around 10000 line of input. Each line contains a single integer n (0<n<2³¹-1). Input is terminated by a line containing a zero.

Output

For each line of input produce one line of output. This line contains one integer followed by one floating point number. The integer number denotes the smallest possible value of L and floating-point number denotes the corresponding value of x. This floating-point number should have eight digits after the decimal point.

Sample Input Output for Sample Input

300

2 0.18798830

2 0.05265302

6 0.25637435

Problemsetter: Shahriar Manzoor

Special Thanks: Arifuzzaman Arif, Sohel Hafiz, Derek Kisman

由题中两个formula可以得到：

ln(n) - ln(1 - x) = L.注意到fabs(x) < 1,得到0 < ln(1 - x) < ln2,所以ln(n) - ln2 < L,于是L取[ln(n) - ln2 + 1]即可（[x]代表取小于x的最大整数）

AC CODE

//Memory: 0 KB		Time: 24 MS
//Language: ANSI C 4.1.2		Result: Accepted

#include <stdio.h>
#include <math.h>

int main()
{
    int n, L;
    double x;
    const double u = log(2);
    while(scanf("%d", &n) && n)
    {
       double m = log(n);
       L = ceil(m - u);
       x = 1- exp(m - L);
       printf("%d %.8lf\n", L, x);
    }
    return 0;
}