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主从、环境搭建、主从配制
查看>>
Mysql主从不同步
查看>>
mysql主从同步及清除信息
查看>>
MySQL主从同步相关-主从多久的延迟?
查看>>
mysql主从同步配置方法和原理
查看>>
mysql主从复制 master和slave配置的参数大全
查看>>
MySQL主从复制几个重要的启动选项
查看>>
MySQL主从复制及排错
查看>>
mysql主从复制及故障修复
查看>>
MySQL主从复制的原理和实践操作
查看>>
webpack loader配置全流程详解
查看>>
mysql主从复制,读写分离,半同步复制实现
查看>>
MySQL主从失败 错误Got fatal error 1236解决方法
查看>>
MySQL主从架构与读写分离实战
查看>>
MySQL主从篇:死磕主从复制中数据同步原理与优化
查看>>
mysql主从配置
查看>>
MySQL之2003-Can‘t connect to MySQL server on ‘localhost‘(10038)的解决办法
查看>>
MySQL之CRUD
查看>>
MySQL之DML
查看>>
Mysql之IN 和 Exists 用法
查看>>