UVA 575-Skew Binary

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A - Skew Binary

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Description

When a number is expressed in decimal, the k-th digit represents a multiple of10^k. (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 of2^k. 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 of2^k⁺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

1012020000000000000000000000000000010100000000000000000000000000000011100111110000011100001011011020000

Sample Output

44214748364632147483647471041110737

题目大意：

给你一个斜二进制的数，叫你算出他的十进制。

题目解析：

这题很简单没什么难度，把字符串转化为数字。再按照题目所给的公式，计算输出就好了。

#include<iostream>#include<string>#include<math.h>using namespace std;int main() {string num;while(cin>>num) {if(num == "0")break;int len = num.size();long long sum = 0;int tmp;for(int i=0;i<len;i++) {tmp = num[i]-'0';sum += tmp * (pow(2,len-i) - 1);}cout<<sum<<endl;}return 0;}

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