• 575 Skew Binary


    When a number is expressed in decimal, the k-th digit represents a multiple of 10k. (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 2k. 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 2k+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 

    44
    2147483646
    3
    2147483647
    4
    7
    1041110737

    思路:把输入的大整数用字符串代替一下,最终的结果不超出int ,用int保存数据。
     1 #include<stdio.h>
     2 #include<string.h>
     3 
     4 int pow(int n)
     5 {
     6     int i,s=1;
     7     for(i=1;i<=n;i++)
     8         s=s*2;
     9     return s;
    10 }
    11 int main()
    12 {
    13     char s[100];
    14     int l,i,num;
    15     while(scanf("%s",s)!=EOF)
    16     {
    17         l=strlen(s);
    18         num=0;
    19         if(s[0]=='0'&&l==1)break;
    20         for(i=l-1;i>=0;i--)
    21             num+=(s[i]-'0')*(pow(l-i)-1);
    22         printf("%d\n",num);
    23     }
    24     return 0;
    25 }
  • 相关阅读:
    [Leetcode]142. Linked List Cycle II
    c程序设计语言笔记001
    把输入一次一个字符复制到输出
    linux笔记
    转 算法之二叉树各种遍历
    将输入复制到输出
    java.lang.OutOfMemoryError处理错误
    字号对照表
    颜色十六进制对照表一
    CSS样式(转)
  • 原文地址:https://www.cnblogs.com/fjutacm/p/2924168.html
Copyright © 2020-2023  润新知