zoj1712 Skew Binary

最新推荐文章于 2018-08-05 21:16:30 发布

原创最新推荐文章于 2018-08-05 21:16:30 发布 · 804 阅读

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本文深入探讨了Skew Binary编码的概念、特点，并提供了将Skew Binary编码转换为十进制数值的方法及示例。通过具体实例演示了如何利用公式计算Skew Binary数对应的十进制值，旨在帮助读者理解这种独特的数制系统。

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

81307(10) = 8 * 10^4 + 1 * 10 ^3 + 3 * 10^2 + 0 * 10^1 + 7 * 10^0
= 80000 + 1000 + 300 + 0 + 7
= 81307.

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

10011(2) = 1 * 2^4 + 0 * 2^3 + 0 * 2^2 + 1 * 2^1 + 1 * 2^0
= 16 + 0 + 0 + 2 + 1
= 19.

In skew binary, the kth 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,

10120(skew) = 1 * (2^5-1) + 0 * (2^4-1) + 1 * (2^3-1) + 2 * (2^2-1) + 0 * (2^1-1)
= 31 + 0 + 7 + 6 + 0
= 44.

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

44
2147483646
3
2147483647
4
7

1041110737

将数字用gets()函数输入数组；

用比较函数strcmp(s,"0") != 0结束输入；

在语句sum=sum+(s[i]-'0')*(pow(2,l)-1);中(s[i]-'0')，因为ASCII码数字与输入数字大小不同！！！！

代码如下：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;

int main(){
    int i;
    double l,sum;
    char s[31];
    while(gets(s)&&strcmp(s,"0") != 0)
    {
       l=strlen(s);
       sum=0;
       for(i=0;i<strlen(s);i++,l--)
          sum=sum+(s[i]-'0')*(pow(2,l)-1);
       cout<<int(sum)<<endl;
    }    
    return 0;
}