UVa 575 - Skew Binary-优快云博客

本文介绍UVa575 SkewBinary问题，解释了偏斜二进制的概念，提供了一种将偏斜二进制转换为十进制的算法实现，并附带示例输入输出。

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UVa 575 - Skew Binary

1 题目
2 思路
3 代码
4 参考

1题目

===================

Skew Binary

When a number is expressed in decimal, thek-th digit represents a multiple of 10^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, thek-th digit represents a multiple of 2^k. For example,

$\begin{displaymath}10011_2 = 1 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 +1 \times 2^0 = 16 + 0 + 0 + 2 + 1 = 19.\end{displaymath}$

Inskew binary, thek-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.)

10120
200000000000000000000000000000
10
1000000000000000000000000000000
11
100
11111000001110000101101102000
0

Sample Output

Miguel A. Revilla
1998-03-10

===================

2思路

题目关键是结果不会超出整形所表示的范围，所以不用按大数运算处理，另外输入可以用字符串表示

3代码

/*
 * Problem: UVa 575 - Skew Binary
 * Lang: ANSI C
 * Time: 0.015
 * Author: minix
 */

#include <stdio.h>
#include <string.h>

#define N 32+5
int base[N];
char input[N];

int main() {
  int i, j;
  int por, len;
  double sum;

  por = 2;
  for (i=1; i<=31; i++) {
    base[i] = por -1;
    por *= 2;
  }

  while (scanf ("%s", input) != EOF) {
    if (!strcmp(input,"0")) break;
    len = strlen (input);
    j = 1; sum = 0;
    for (i=len-1; i>=0; i--)
      sum += base[j++] * (input[i]-'0');
    printf ("%.0lf\n", sum);
  }

  return 0;
}