575 - Skew Binary

最新推荐文章于 2019-09-08 21:23:30 发布

SIO__Five

最新推荐文章于 2019-09-08 21:23:30 发布

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本文介绍了一种特殊的进制——偏斜二进制，并提供了一个C++程序用于将偏斜二进制数转换为十进制数。偏斜二进制中，k位上的值代表2^(k+1)-1，且最低位可以为2。

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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 2³¹ - 1 = 2147483647.

Sample Input

10120
200000000000000000000000000000
10
1000000000000000000000000000000
11
100
11111000001110000101101102000
0

Sample Output

给你一个数，是按照一个特殊的进制规格的。其中k位上代表了2^（k+1) -1 。然后就是求出十进制。水题

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int main ()
{
    char skew[50];
    int sum,i;
    while(cin>>skew)
    {
        sum=0;
        if (strcmp(skew,"0")==0) break;
        int len=strlen(skew);
        for (i=len-1; i>=0; i--)
            sum+=(skew[i]-'0')*(pow(2,len-i)-1);
        cout<<sum<<endl;
    }
    return 0;
}