codeforces B. Spreadsheets 计算excel的队列

codeforces B. Spreadsheets 计算excel的行列

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.

Sample test(s)

input
2
R23C55
BC23

output
BC23
R23C55

没想到这道题看起来这么简单，要写出无Bug的程序居然这么麻烦，要处理挺多情况的。

如：

1 判断是什么格式，然后转到相应的转换

2 26进制和10进制互相转换，如何处理

3 26进制还不是从0开始的，要如何处理

对应这里分了四个函数，应该都很清晰了。

#pragma once
#include <assert.h>
#include <math.h>
#include <algorithm>
#include <utility>
#include <queue>
#include <deque>
#include <vector>
#include <string>
#include <iostream>
using namespace std;

string RCtoABC(string &rc)
{
	string num;
	unsigned i = 1;
	for ( ; i < rc.size() && rc[i] != 'C'; i++)
	{
		num.push_back(rc[i]);
	}
	string abc;
	int a = 0;
	for (i++; i < rc.size(); i++)
	{
		a = a * 10 + rc[i] - '0';
	}
	while (a)
	{
		int b = (a - 1) % 26;//注意这里的转换方法
		a = (a - 1) / 26;
		abc.push_back(b + 'A');
	}
	reverse(abc.begin(), abc.end());
	return abc+num;
}

pair<int, string> ABCtoRC(string &abc)
{
	unsigned i = 0;
	int rnum = 0;
	for ( ; i < abc.size(); i++)
	{
		if (abc[i] < 'A' || 'Z' < abc[i]) break;
		rnum = rnum * 26 + (abc[i] - 'A' + 1);//注意+1
	}
	return pair<int, string>(rnum, abc.substr(i));
}

bool isRC(string &s)
{
	if (s[0] != 'R') return false;
	unsigned i = 1;
	for ( ; i < s.size() && 'A' <= s[i] && s[i] <= 'Z'; i++);//注意判断，如RC12
	for ( ; i < s.size() && '0' <= s[i] && s[i] <= '9'; i++);
	if (i >= s.size() || s[i] != 'C') return false;
	return true;
}

void Spreadsheets()
{
	int n;
	cin>>n;
	string s;
	while (n--)
	{
		cin>>s;
		if (isRC(s))
		{
			cout<<RCtoABC(s)<<endl;
		}
		else
		{
			pair<int, string> pis = ABCtoRC(s);
			cout<<'R'<<pis.second<<'C'<<pis.first<<endl;
		}
	}
}

codeforces B. Spreadsheets 计算excel的队列

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