skew二进制算法

 

skew二进制

Time Limit 1000ms

Memory Limit 65536K

description

When a number is expressed in decimal, the kth digit represents a multiple of 10k. (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 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

#include <iostream>

#include "cmath"
using namespace std;


int main()
{
	int a,ax[100];char t[100];int n,i,j;
	
	
	while(scanf("%s",t)!=EOF)
	{a=0;
		n=strlen(t);
		if(n==1&&t[0]=='0')
			break;
		
		for(i=n-1;i>=0;i--)
		{
			a=a+(t[i]-'0')*(pow(2.0,n-i)-1);}
		
		cout<<a<<endl;
	}

}