数制形式转换问题

问题：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.)

Sample Input
10120
200000000000000000000000000000
10
1000000000000000000000000000000
11
100
11111000001110000101101102000
0

Sample Output
44
2147483646
3
2147483647
4
7
1041110737

回答：主要是pow函数使用。

#include <iostream>  
#include <string>  
#include <cmath>  
using namespace std;  
int main(){  
    string s;  
    while(1){  
        cin>>s;  
        if (s == "0")  
            break;  
        long result = 0;  
        int k = 1;  
        string::iterator it;  
        for (it = s.end()-1;it!=s.begin();it--)  
        {  
            result += (*it - '0')*(pow(double(2),k++) - 1);  
        }  
        result += (*it - '0')*(pow(double(2),k) - 1);  
        cout<<result<<endl;  
    }  
    return 0;  
}