UVA 11809 Floating-Point Numbers（暴力）

最新推荐文章于 2022-01-11 15:16:22 发布

Shili_Xu

最新推荐文章于 2022-01-11 15:16:22 发布

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分类专栏：暴力文章标签： UVA 暴力

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本文探讨了如何根据浮点数能表示的最大值来确定其尾数位数(M)和指数位数(E)的方法。通过解析输入的浮点数最大值，采用科学记数法转换并对比预设表格，实现了快速准确地找出浮点数格式。

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Floating-point numbers arerepresented differently in computers than integers. That is why a 32-bitfloating-point number can represent values in the magnitude of 10³⁸whilea 32-bit integer can only represent values as high as 2³².

Although thereare variations in the ways floating-point numbers are stored in Computers, inthis problem we will assume that floating-point numbers are stored in thefollowing way:

Floating-pointnumbers have two parts mantissa and exponent. M-bits are allotted for mantissa and E bits are allotted for exponent. There is also one bit thatdenotes the sign of number (If this bit is 0 then the number is positive and ifit is 1 then the number is negative) and another bit that denotes the sign ofexponent (If this bit is 0 then exponent is positive otherwise negative). Thevalue of mantissa and exponent together make the value of the floating-pointnumber. If the value of mantissa is m thenit maintains the constraints 1. The left most digit ofmantissa must always be 1 to maintain the constraint 1. So this bit is not stored asit is always 1. So the bits in mantissa actually denote the digits at the rightside of decimal point of a binary number (Excluding the digit just to the rightof decimal point)

In the figure above we can see a floating-point number where M = 8 and E = 6. The largest value this floating-point number can representis (in binary) 0.111111111₂×2¹¹¹¹¹¹². The decimal equivalent to this number is: 0.998046875 × 2⁶³= 9205357638345293824₁₀. Giventhe maximum possible value represented by a certain floating point type, youwill have to find how many bits are allotted for mantissa (M) and how many bits are allotted for exponent (E) in that certain type.

Input

The input filecontains around 300 line of input. Each line contains a floating-point number F that denotes the maximum value thatcan be represented by a certain floating-point type. The floating point numberis expressed in decimal exponent format. So a number AeB actually denotes the value A×10^B. A line containing ‘0e0’ terminates input. The value of A will satisfy the constraint 0 < A < 10 and will have exactly 15digits after the decimal point.

Output

For each line ofinput produce one line of output. This line contains the value of M and E. You can assume that each of the inputs (except the last one) hasa possible and unique solution. You can also assume that inputs will be suchthat the value of M and E will follow the constraints: 9 ≥ M ≥ 0 and 30 ≥ E ≥ 1. Also there is no need to assume that (M + E + 2) will be amultiple of 8.

Sample Input

5.699141892149156e76

9.205357638345294e18

0e0

Sample Output

5 8

8 6

【思路】

题目给定一个最大的浮点数，用AeB表示（科学计数法），求需要的尾数尾数M和阶数尾数E。依据十进制和二进制两种表示方式的等价关系，取对数打表就好了。然后利用流读入查表。

【代码】

#include <iostream>
#include <cstring>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;

const int MAXM = 15, MAXE = 35;
const double EPS = 1e-5;

double res[MAXM][MAXE];

int main()
{
    for (int i = 0; i <= 9; i++)
        for (int j = 1; j <= 30; j++) {
            double m = 1.0 - pow(2, -(i + 1)), e = pow(2, j) - 1;
            res[i][j] = log(m) + e * log(2);
        }
    string buff;
    stringstream stream;
    while (cin >> buff && buff != "0e0") {
        for (int i = 0; i < buff.length(); i++)
            if (buff[i] == 'e') buff[i] = ' ';
        stream.str("");stream.clear();
        stream << buff;
        double a;
        int b;
        stream >> a >> b;
        for (int i = 0; i <= 9; i++)
            for (int j = 1; j <= 30; j++) {
                double ans = log(a) + b * log(10);
                if (-EPS < ans - res[i][j] && ans - res[i][j] < EPS) {
                    cout << i << " " << j << endl;
                    break;
                }
            }
    }
    return 0;
}