UVA - 575 - Skew Binary (简单数论!)
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Description
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,
When a number is expressed in binary, the k-th digit represents a multiple of 2k. For example,
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,
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
1012020000000000000000000000000000010100000000000000000000000000000011100111110000011100001011011020000
Sample Output
44214748364632147483647471041110737
Miguel A. Revilla
1998-03-10
Source
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Mathematics :: Ad Hoc Mathematics Problems :: Base Number Variants
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 1. Elementary Problem Solving :: Maths - Number Theory
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Mathematics :: Ad Hoc Mathematics Problems :: Base Number Variants
AC代码:
#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define LL long longusing namespace std;int main(){char a[40];while(scanf("%s", a)!=EOF){if(strlen(a)==1 && a[0] == '0')break;int len = strlen(a);LL bei=2, ans = 0;for(int i=len-1; i>=0; i--){ans += ( (bei-1) * (a[i]-'0') );bei*=2;}printf("%lld\n", ans);}return 0;}
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