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When a number is expressed in decimal, the k-th digit represents a multiple of 10k
. (Digits are numbered
from right to left, where the least significant digit is number 0.) For example,
8130710 = 8 × 104 + 1 × 103 + 3 × 102 + 0 × 101 + 7 × 100 = 80000 + 1000 + 300 + 0 + 7 = 81307.
When a number is expressed in binary, the k-th digit represents a multiple of 2
k
. For example,
100112 = 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 k-th 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,
10120skew = 1×(25 −1)+ 0×(24 −1)+ 1×(23 −1)+ 2×(22 −1)+ 0×(21 −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 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 2
31 − 1 =
2147483647.
Sample Input
10120
200000000000000000000000000000
10
1000000000000000000000000000000
11
100
11111000001110000101101102000
0
Sample Output
44
2147483646
3
2147483647
4
7
1041110737

这里有个细节问题，但还是个水题.

#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
    int sum=0;
    int i[32];
    i[0]=1;
    for(int a = 1;a < 32;a ++)
        i[a]=i[a-1]*2;
    char c[1005];
    while(~scanf("%s",c)&&c[0]!='0')
    {
        sum=0;
        for(int a = 0;a < strlen(c);a ++)
            sum+=(c[a]-48)*(i[strlen(c)-a]-1);
         printf("%d\n",sum);
    }
    return 0;
}

本文标签： 数论简单uvabinarySkew

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