Baiti 2973 Skew data solution report

Link: http://poj.grids.cn/practice/2973/

Question:

Total time limit:

1000 ms

Memory limit:

65536kB

Description

In the skew binary representation, the k-bit value xk indicates xk * (2 k + 1-1 ). The possible numbers on each bit are 0 or 1, and the last non-zero bit can be 2, for example, 10120 (skew) = 1 * (25-1) + 0 * (24-1) + 1 * (23-1) + 2 * (22-1) + 0 * (21-1) = 31 + 0 + 7 + 6 + 0 = 44. the top 10 skew numbers are 0, 1, 2, 10, 11, 12, 20, 100, 101, and 102.

Input

The input contains one or more rows. Each row contains an integer n. If n = 0, the input ends. Otherwise, n is a skew.

Output

For each input, it is output in decimal format. After being converted to decimal, n cannot exceed 231-1 = 2147483647

Sample Input

1012020000000000000000000000000000010100000000000000000000000000000011100111110000011100001011011020000

Sample output

44214748364632147483647471041110737

Code:

 #include <iostream> #include <cstdlib> #include <cstdio> #include <cstring>     MAX =                        (scanf(,chs) != EOF && strcmp(chs,          sum =          step =          ( i = strlen(chs)-; i >= ; i--              sum += (chs[i] - ) * (step -              step *=                    cout<<sum<<        }

Ideas:

1. Use int to save the result.

2. You must use a string to read data.

3. Use step for Optimization