UVA - 575 - Skew Binary （简单数论！）

UVA - 575 Skew Binary Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu Submit Status 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 10120 200000000000000000000000000000 10 1000000000000000000000000000000 11 100 11111000001110000101101102000 0 Sample Output 44 2147483646 3 2147483647 4 7 1041110737 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

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define LL long long
using 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;
}