Skew Binary 1565 (简单数学)

Skew Binary

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 10717 Accepted: 6838

Description

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 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

#include<stdio.h>#include<string.h>#include<math.h>int main(){int a[1001];char s[1001];int n,sum,i,j,l;while(scanf("%s",s)!=EOF){sum=0;memset(a,0,sizeof(a));if(s[0]=='0')    break; l=strlen(s); for(i=0;i<l;i++) a[i]=s[i]-'0';for(i=0;i<l;i++) sum+=a[i]*(pow(2,l-i)-1); printf("%d\n",sum);}return 0;}