Skew Binary——进制转换

最新推荐文章于 2020-06-03 17:38:36 发布

原创最新推荐文章于 2020-06-03 17:38:36 发布 · 708 阅读

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本文介绍了SkewBinary进制的定义、特点及其在特定应用中的优势。详细解释了如何将SkewBinary数转换为十进制数，并提供了处理超出整型范围的策略。

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Skew Binary

When a number is expressed in decimal, the k-th digit represents a multiple of 10^k. (Digits are numbered from right to left, where the least significant digit is number 0.) For example,

$\begin{displaymath}81307_{10} = 8 \times 10^4 + 1 \times 10^3 + 3 \times 10^2 + ......mes 10^1 +7 \times 10 0 = 80000 + 1000 + 300 + 0 + 7= 81307.\end{displaymath}$

When a number is expressed in binary, the k-th digit represents a multiple of 2^k. For example,

$\begin{displaymath}10011_2 = 1 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 +1 \times 2^0 = 16 + 0 + 0 + 2 + 1 = 19.\end{displaymath}$

In skew binary, the k-th digit represents a multiple of 2^k+1 - 1. The only possible digits are 0 and 1, except that the least-significant nonzero digit can be a 2. For example,

$\begin{displaymath}10120_{skew} = 1 \times (2^5 - 1) + 0 \times (2^4-1) + 1 \tim......2 \times (2^2-1) + 0 \times (2^1-1)= 31 + 0 + 7 + 6 + 0 = 44.\end{displaymath}$

The first 10 numbers in skew binary are 0, 1, 2, 10, 11, 12, 20, 100, 101, and 102. (Skew binary is useful in some applications because it is possible to add 1 with at most one carry. However, this has nothing to do with the current problem.)

Input

The input file contains one or more lines, each of which contains an integer n . If n = 0 it signals the end of the input, and otherwise n is a nonnegative integer in skew binary.

Output

For each number, output the decimal equivalent. The decimal value of n will be at most 2 ³¹ - 1 = 2147483647.

Sample Input

10120
200000000000000000000000000000
10
1000000000000000000000000000000
11
100
11111000001110000101101102000
0

Sample Output

给出二进制，按照 $\begin{displaymath}10120_{skew} = 1 \times (2^5 - 1) + 0 \times (2^4-1) + 1 \tim......2 \times (2^2-1) + 0 \times (2^1-1)= 31 + 0 + 7 + 6 + 0 = 44.\end{displaymath}$ 的规则进行进制转换，如果得出的结果超出int范围，打印2147483647

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

int main()
{
    char cunc[1000];
    int l,i;
    long long jg,num;
    memset(cunc,0,sizeof(cunc));
    while(scanf("%s",cunc))
    {
        l = strlen(cunc);
        if(strcmp(cunc,"0") == 0)
            break;
        if(l >= 31)
        {
            printf("2147483647\n");
            continue;
        }
        jg = 0;
        for(i = 0;i < l;i++)
        {
            num = pow(2,l-i);
            jg = jg + (cunc[i] - '0') * (num-1);
        }
        if(jg >= 2147483648)
            printf("2147483647\n");
        else
            printf("%lld\n",jg);
    }
    return 0;
}