博客
关于我
B. Spreadsheets(进制转换,数学)
阅读量:426 次
发布时间:2019-03-06

本文共 2421 字,大约阅读时间需要 8 分钟。

B. Spreadsheets
time limit per test
10 seconds
memory limit per test
64 megabytes
input
standard input
output
standard output

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.

Input

The first line of the input contains integer number n (1 ≤ n ≤ 105), 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 106 .

Output

Write n lines, each line should contain a cell coordinates in the other numeration system.

Examples
input
2 R23C55 BC23
output
BC23 R23C55

 

 

 

#include 
#include
#define M 1*10^6+10char str[M];/* 十进制变为二十六进制 */void solve_10_to_26(char str[]){ char re[M]; int p = M - 1; int r, c; sscanf(str, "R%dC%d", &r, &c); re[p--] = 0; while(r) { re[p--] = r % 10 + '0'; r = r/10; } while(c) { re[p--] = (c-1) % 26 + 'A'; c = (c-1)/26; } printf("%s\n", &re[p+1]);}/* 二十六进制变为十进制 */void solve_26_to_10( char str[] ){ char cc[M]; int c = 0; int r; sscanf( str, "%[A-Z]%d", &cc, &r ); int len = strlen(cc); int p = 0; while(p < len) { c = c + cc[p++] - 'A' + 1; c = c * 26; } c = c / 26; printf("R%dC%d\n", r, c);}int main(){ int n; scanf("%d", &n); while(n--) { scanf("%s", &str); int a, b; if ( sscanf(str, "R%dC%d", &a, &b ) == 2) solve_10_to_26(str); else solve_26_to_10(str); } return 0;}
View Code

 

转载地址:http://gltuz.baihongyu.com/

你可能感兴趣的文章
mysql函数汇总之系统信息函数
查看>>
MySQL函数简介
查看>>
mysql函数遍历json数组
查看>>
MySQL函数(转发)
查看>>
mysql分区表
查看>>
MySQL分层架构与运行机制详解
查看>>
mysql分库分表中间件简书_MySQL分库分表
查看>>
MySQL分库分表会带来哪些问题?分库分表问题
查看>>
MySQL分组函数
查看>>
MySQL分组查询
查看>>
Mysql分表后同结构不同名称表之间复制数据以及Update语句只更新日期加减不更改时间
查看>>
mySql分页Iimit优化
查看>>
MySQL分页查询
查看>>
WebDriverException:未知错误:对于旧版本的 Google Chrome,在 Python 中找不到带有 Selenium 的 Chrome 二进制错误
查看>>
mysql列转行函数是什么
查看>>
mysql创建函数报错_mysql在创建存储函数时报错
查看>>