Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Floating point numbers are represented at the hardware level of most computer systems in normalized binary scientific notation. We will consider a hypothetical computer in which floating point numbers are represented in a 16-bit word. An example is shown in Figure 1:

In the common Intel chips, the exponent of a float has 8 bits and the mantissa has 23 bits, otherwise the representation is the same as in Figure 1.

We will show that this particular example represents the number whose usual decimal representation is -10.375.

The leftmost bit (the sign bit) is 1 if the floating point number is negative, 0 otherwise.

The next seven bits (1000010 in this example) represent the Exponent component of the normalized binary scientific notation. The following two operations will reveal the exponent's actual value:

Evaluate the binary number 1000010: it is 1*64+0*32+0*16+0*8+0*4+1*2+0*1 = 66.

Subtract the bias correction 63 from 66, which yields 3 as the actual value of the exponent that we will use below (the purpose of the bias correction is to allow negative exponents. As long as the exponent has seven bits, which will always be the case in this program, the bias correction is 63 = 2^7-1 - 1).

The remaining 8 bits (01001100 in this example) constitute the mantissa, which is the fractional part of the normalized binary scientific notation. You will see in the next step how it is used.

The sign bit, the mantissa, and the exponent are used to calculate the value of the floating point number:

The period preceding the mantissa is the "binary point" which has the same meaning as the decimal point in a decimal (base ten) representation.

In the normalized binary scientific notation there is only one binary digit preceding the binary point, and it is always 1 (except if the floating point number is the number zero). It is called the characteristic. Since it is always 1, it does not explicitly appear in the 16-bit representation of Figure 1.

Since the number 1.01001100 is in binary, the digits to the right of the binary point represent successive negative powers of 2. Therefore, the number shown in Figure 2 can be re-written as:

Here the result has been expressed using scientific notation (base ten). Most programming languages would output this result as -1.0375e+001. The use of a lower case "e" or upper case "E", and the number of leading zeros in the exponent can vary from one programming language to another.

An important exception: if all bits except the sign bit are zero, then the floating point number is zero, regardless of the sign bit.

Input 

The input contains an undetermined number of lines. Each line contains, starting in column 1, exactly 16 printable characters which can be any permutation of 0s and 1s. The first character represents the sign bit, the next seven characters, the exponent, and the last eight characters, the mantissa of a floating point number in normalized binary scientific notation according to the conventions discussed on the preceding page.

Output 

Each floating point number read from the input will be written by your program to the output file, one number per line.

The following formatting conventions are required in your output:

The floating point numbers will be displayed in scientific notation (base ten).

Column 1 will either be blank (indicating a positive number or zero) or contain a minus sign (indicating a negative number).

Column 2 will contain a digit.

Column 3 will contain the decimal point.

Columns 4 - 9 will contain six additional significant digits of the floating point number.

The remaining columns will specify the exponent of scientific notation. The sign and magnitude of the exponent must of course be correct, but the exact format in which the exponent is displayed will be determined by the programming language you use.

Sample Input 

1100001001001100
0011111100000000
1011111110000000
0000000010101010
0011011111100000
1001111011100000
0101011001010101
0100011011101101
0111111111111111
1100001000101100
0000000000000000
1000000000000000

Sample Output 

Program 6 by team X
-1.037500e+001
 1.000000e+000
-1.500000e+000
 1.804180e-019
 7.324219e-003
-2.182787e-010
 1.117389e+007
 2.465000e+002
 3.682143e+019
-9.375000e+000
 0.000000e+000
 0.000000e+000
End of program 6 by team X

---

遇見最詭的一道題目，一開始想說用 %e 自動輸出，但是沒過。

後來改成 double 的手動格式處理，也沒有過。

最後使用 float，居然過了。<(_ _)>

#include <stdio.h>
#include <math.h>
int main() {
    puts("Program 6 by team X");
    char s[105];
    while(scanf("%s", s) == 1) {
        int zero = 1, i, j;
        for(i = 1; i < 16; i++)
            zero &= s[i] == '0';
        if(zero) {
            puts(" 0.000000e+000");
            continue;
        }
        long long e = 0, f = 0;
        for(i = 1; i <= 7; i++)
            e = (e<<1) + s[i]-'0';
        e = e-63;
        for(i = 8; i < 16; i++)
            f = (f<<1)|(s[i]-'0');
        double ret = (1+f/256.0)*pow(2, e);
        e = 0;
        if(ret > 10-1e-6) {
            while(ret > 10-1e-6) {
                e++;
                ret /= 10;
            }
        } else if(ret < 1 + 1e-6) {
            while(ret < 1+1e-6) {
                e--;
                ret *= 10;
            }
        }
        if(s[0] == '1') printf("-");
        else            printf(" ");
        printf("%.6fe", ret);
        if(e >= 0)  printf("+%03d\n", e);
        else        printf("-%03d\n", -e);
    }
    puts("End of program 6 by team X");
    return 0;
}