Codeforces Beta Round #1 B. Spreadsheets 模拟

B. Spreadsheets

题目连接：

http://www.codeforces.com/contest/1/problem/B

Description

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 Input

2
R23C55
BC23

Sample Output

BC23
R23C55

Hint

题意

表格有两种表示方法，第一种

比如R23C55，就表示第23行，55列

第二种：

比如BC23，就表示在第BC列，23行，BC是一个26进制数，A是1，Z是26，BC就表示55=2*26+3

然后给你其中一种，让你转化成另外一种

题解：

模拟题，瞎跑跑就好了……

简单模拟题，进制转换，直接看(这个数-1)%26就好了。

代码

#include<bits/stdc++.h>
using namespace std;

string s;
void solve1()
{
    int R=0,C=0;
    int flag = 0;
    for(int i=1;i<s.size();i++)
    {
        if(s[i]=='C')flag=1;
        if(s[i]=='C')continue;
        if(flag==0)R=R*10+(s[i]-'0');
        else C=C*10+(s[i]-'0');
    }
    string ans;
    while(C)
    {
        int p = (C-1)%26;
        C=(C-1)/26;
        ans+=(p+'A');
    }
    reverse(ans.begin(),ans.end());
    cout<<ans<<R<<endl;
}
void solve2()
{
    int flag = 0;
    int C=0;
    for(int i=0;i<s.size();i++)
    {
        if(s[i]<='9'&&s[i]>='0'&&flag==0)
        {
            flag = 1;
            cout<<"R";
        }
        if(flag==1)cout<<s[i];
        else
            C=C*26+(s[i]-'A'+1);
    }
    cout<<"C"<<C<<endl;
}
int main()
{
    int time;
    scanf("%d",&time);
    while(time--)
    {
        cin>>s;
        int flag1=0,flag2=0,flag3=0;
        for(int i=0;i<s.size();i++)
        {
            if(s[i]=='R')flag1++;
            if(s[i]=='C')flag2++;
            if(s[i]<='Z'&&s[i]>='A')
                flag3++;
        }
        if(s[0]=='R'&&s[1]=='C')flag1=0;
        if(flag3==2&&flag1&&flag2)
            solve1();
        else
            solve2();
    }
}