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用户以及权限的管理。
查看>>
MySQL用户权限配置:精细控制和远程访问的艺术!------文章最后有惊喜哦。
查看>>
mysql用户管理、常用语句、数据分备份恢复
查看>>
MySQL留疑问:left join时选on还是where?
查看>>
mysql登陆慢问题解决
查看>>
MySQL的 DDL和DML和DQL的基本语法
查看>>
mysql的 if else , case when then, IFNULL
查看>>
MySQL的10种常用数据类型
查看>>
MySQL的btree索引和hash索引的区别
查看>>
mysql的cast函数
查看>>
MySql的CRUD(增、删、改、查)操作
查看>>
MySQL的DATE_FORMAT()函数将Date转为字符串
查看>>
mysql的decimal与Java的BigDecimal用法
查看>>
MySql的Delete、Truncate、Drop分析
查看>>
MySQL的Geometry数据处理之WKB方案
查看>>
MySQL的Geometry数据处理之WKT方案
查看>>
mysql的grant用法
查看>>
Mysql的InnoDB引擎的表锁与行锁
查看>>
mysql的InnoDB引擎索引为什么使用B+Tree
查看>>
MySQL的InnoDB默认隔离级别为 Repeatable read(可重复读)为啥能解决幻读问题?
查看>>