UVA 11809 - Floating-Point Numbers【浮点数】

11809 - Floating-Point Numbers

Time limit: 1.000 seconds

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=226&page=show_problem&problem=2909

Floating-point numbers are represented differently in computers than integers. That is why a 32-bitfloating-point number can represent values in the magnitude of 10^38 while a 32-bit integer can onlyrepresent values as high as 2^32.

Although there are variations in the ways floating-point numbers are stored in Computers, in thisproblem we will assume that floating-point numbers are stored in the following way:

Floating-point numbers have two parts mantissa and exponent. M-bits are allotted for mantissaand E bits are allotted for exponent. There is also one bit that denotes the sign of number (If thisbit is 0 then the number is positive and if it is 1 then the number is negative) and another bit thatdenotes the sign of exponent (If this bit is 0 then exponent is positive otherwise negative). The value ofmantissa and exponent together make the value of the floating-point number. If the value of mantissais m then it maintains the constraints 12 ≤ m < 1. The left most digit of mantissa must always be 1 tomaintain the constraint 12 ≤ m < 1. So this bit is not stored as it is always 1. So the bits in mantissaactually denote the digits at the right side of decimal point of a binary number (Excluding the digitjust to the right of decimal point)

In the figure above we can see a floating-point number where M = 8 and E = 6. The largest valuethis floating-point number can represent is (in binary)

0.111111111(2)×2^111111(2). The decimal equivalentto this number is: 0.998046875 × 2^63 = 9205357638345293824(10). Given the maximum possible valuerepresented by a certain floating point type, you will have to find how many bits are allotted formantissa (M) and how many bits are allotted for exponent (E) in that certain type.

Input

The input file contains around 300 line of input. Each line contains a floating-point number F thatdenotes the maximum value that can be represented by a certain floating-point type. The floating pointnumber is 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 andwill have exactly 15 digits after the decimal point.

Output

For each line of input produce one line of output. This line contains the value of M and E. You canassume that each of the inputs (except the last one) has a possible and unique solution. You can alsoassume that inputs will be such that the value of M and E will follow the constraints: 9 ≥ M ≥ 0 and30 ≥ E ≥ 1. Also there is no need to assume that (M + E + 2) will be a multiple of 8.

Sample Input

5.699141892149156e76
9.205357638345294e18
0e0

Sample Output

5 8
8 6

题目大意就是说，给出一个浮点数，求出尾数和阶码所需的最小位数。反正这道题是看了人家题解看了好久才懂的。开二维数组M[ ][ ] 和E[ ][ ] 分别记录对应位数的尾数和阶码的组合时尾数和指数所能表示的最大值，然后打表查找比较。对于一个a位二进制数的尾数，它的十进制尾数值m=1-2^(-a-1) 对于一个b位二进制数的阶码，可表示的指数为2^e=2^[ 2 ^(b-1)] 然后把这个数转换成对应的以10为底的计数法表示出来，m*2^e=c*10^d，两边取对数，得到log10(m)+e*log10(2)=log10(c)+d，令左边为t ，则 d=t-log10(c)，由于1<=c<=10，所以log10(c)<1，d为指数，为正，所以d=t/1(自动向下取整），所以c=10^(t-d)，这样打表。。。

我看的是：http://www.th7.cn/Program/cp/201412/334378.shtml

#include 
#include
#include
#include
using namespace std;
char str[110],t[110];
double M[10][31];
long long E[10][31];
void dabiao()
{
    for(int a=0;a<10;++a)
    {
        for(int b=1;b<31;++b)
        {
            double m=1.0-1.0/(1<<(a+1));
            long long e=(1<