CodeForces 1B-字符串，进制转换与数学

一个萌新的成长之路

Background

• 同学们都回家了，只有我和wjh还有邢神在机房敲代码，吃random口味的方便面……

Description

Translated by @PC_DOS from luogu

• In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc.
• The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23.
• Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example.
• Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system.
• 人们常用的电子表格软件(比如: Excel)采用如下所述的坐标系统:
• 第一列被标为A，第二列为B，以此类推，第26列为Z。接下来为由两个字母构成的列号: 第27列为AA，第28列为AB...在标为ZZ的列之后则由三个字母构成列号，如此类推。
  行号为从1开始的整数。
• 单元格的坐标由列号和行号连接而成。比如，BC23表示位于第55列23行的单元格。
  有时也会采用被称为RXCY的坐标系统，其中X与Y为整数，坐标(X,Y)直接描述了对应单元格的位置。比如，R23C55即为前面所述的单元格。
• 您的任务是编写一个程序，将所给的单元格坐标转换为另一种坐标系统下面的形式。

Input&Output

Input Format：

• The first line of the input contains integer number n (1<=n<=10^6) , the number of coordinates in the test. Then there follow n lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 10^6.
• 第一行一个整数n(1<=n<=10^5)，表示将会输入的坐标的数量。
• 接下来n行，每行一个坐标。
• 注意: 每个坐标都是正确的。此外不会出现行号或列号大于10^6的单元格。

Output Format:

• Write n lines, each line should contain a cell coordinates in the other numeration system.
• n行，每行一个被转换的坐标。

Sample

Input:

2
R23C55
BC23

Output:

BC23
R23C55

Solution

• 本题并不难，涉及字符串处理和一点数学.
  首先是判断输入数据属于哪一种类型，由于数据规模较小，本蒟蒻选择了暴力.
  这道题其实相当于一个进制互换，而我们发现行数R是不需要计算的，也就是需要考虑列数C的转化.
• 不难发现对于一个字符串“C(1)C(2)C(3)……C(n)”，给定每个字符的编号为i(A=1，B=2……)，她所代表的列数即为i(n)×26^0+i(n-1)×26^1+i(n-2)×26^2.
• 数字转字符串呢？类比其他进制转换的过程，我们可以知道每次将c对26取模，结果就是对应字符串的第n，n-1，n-2……位.进行每次操作后，c减去 c%26，再用26除即可.
• 需要注意的是，26%26=0，需要对Z进行特判，作差时也要注意减去26.

• 代码如下：

  #include<iostream>
  #include<cstdio>
  #include<cstring>
  using namespace std;
  char s[1000];
  char alphabet[27] = { 'Z','A','B','C','D','E',
  'F','G','H','I','J','K',
  'L','M','N','O','P','Q',
  'R','S','T','U','V','W',
  'X','Y','\0' };
  inline bool check(char *s)//检查类型
  {
  int len=strlen(s);
  int type=0;
  for(int i=0;i<len;++i)
  {
      if(s[i]>='0'&&s[i]<='9')type=1;
      if(type&&s[i]=='C')return true;
  }
  return false;
  }
  inline void rd()
  {
  int r = 0, c = 0;
  int bit = 1;
  scanf("%s", &s);
  char output[1000];
  if (check(s))
  {
      int i = 1;
      while (s[i] >= '0'&&s[i] <= '9')
      {
          r = (r << 1) + (r << 3) + (s[i] ^ 48);//类似于读入优化，提速の小技巧
          i++;
      }
      i++;
      while (s[i] >= '0'&&s[i] <= '9')
      {
          c = (c << 1) + (c << 3) + (s[i] ^ 48);
          i++;
      }
      while (c != 0)//特判
      {
          int t=c%26;
          output[bit++] = alphabet[t];
          c -= t?t:26;
          c /= 26;
      }
      for (int j = bit-1;j >=1;--j)printf("%c", output[j]);
      printf("%d\n", r);
      return;
  }
  else {
      int i = 0;
      r=0;
      c=0;
      while (s[i] >= 'A'&&s[i] <= 'Z')
      {
          c = (c << 1) + (c << 3) + (c << 4) + s[i] + 1 - 'A';
          i++;
      }
      while (s[i] >= '0'&&s[i] <= '9')
      {
          r = (r << 1) + (r << 3) + (s[i] ^ 48);
          i++;
      }
      printf("R%dC%d\n", r, c);
      return;
  }
  }
  int main()
  {
  int n;
  scanf("%d", &n);
  for (int i = 1;i <= n;++i)rd();
  return 0;
  }

• Feb,06,2018 Tue