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  • ZOJ Problem Set–1712 Skew Binary

    Time Limit: 2 Seconds      Memory Limit: 65536 KB


    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


    Source: Mid-Central USA 1997

    #include<iostream>
    
    #include<cmath>
    
    #include<string>
    
    using namespace std;
    
    int main()
    
    {
    
      string drew;
    
      while(cin>>drew && drew != "0")
    
      {
    
        int result = 0;
    
        for(int i = 0; i < drew.length();i++)
    
        {
    
          result += (pow(2, (drew.length() - i)*1.0) - 1)*(drew[i] - '0');
    
        }
    
        cout<<result<<endl;
    
      }
    
      return 0;
    
    }
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  • 原文地址:https://www.cnblogs.com/malloc/p/2482925.html
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