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
nginx日志分割并定期删除
查看>>
Nginx日志分析系统---ElasticStack(ELK)工作笔记001
查看>>
Nginx映射本地json文件,配置解决浏览器跨域问题,提供前端get请求模拟数据
查看>>
nginx最最最详细教程来了
查看>>
Nginx服务器---正向代理
查看>>
Nginx服务器上安装SSL证书
查看>>
Nginx服务器的安装
查看>>
Nginx模块 ngx_http_limit_conn_module 限制连接数
查看>>
nginx添加模块与https支持
查看>>
Nginx用户认证
查看>>
Nginx的location匹配规则的关键问题详解
查看>>
Nginx的Rewrite正则表达式,匹配非某单词
查看>>
Nginx的使用总结(一)
查看>>
Nginx的使用总结(三)
查看>>
Nginx的使用总结(二)
查看>>
Nginx的可视化神器nginx-gui的下载配置和使用
查看>>
Nginx的是什么?干什么用的?
查看>>
Nginx访问控制_登陆权限的控制(http_auth_basic_module)
查看>>
nginx负载均衡和反相代理的配置
查看>>
nginx负载均衡器处理session共享的几种方法(转)
查看>>