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.

The first line of the input contains integer number n (1 ≤ n ≤ 10⁵), the number of coordinates in the test. Then there follow n

Write n lines, each line should contain a cell coordinates in the other numeration system.

2
R23C55
BC23

BC23
R23C55

題目意思：
簡單來說就是10進位制和26進位制的相互轉換，給出RXCY轉換長YX，其中Y為26進位制，模擬題目，不難但是各種陷阱
各種多，看自己的人品了，我都不知道wa了多少次。。。。，還得注意判斷是那種轉換，是RXCY-->XY?還是
XY—>RXCY

AC程式碼：

/**
  *@xiaoran
  *題意意思，26進位制的轉換
  */
#include<iostream>
#include<cstdio>
#include<map>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<cstdlib>
#include<cctype>
#include<cmath>
#define LL long long
using namespace std;
void IntToSta(char *s){
    int sr,sc,len;
    sr=sc=0;
    len=strlen(s);
    for(int i=0;i<len;i++){
        if(isalpha(s[i])){//是字母
            sc=sc*26+(s[i]-'A'+1);
        }
        if(isdigit(s[i])){//是數字
            sr=sr*10+(s[i]-'0');
        }
    }
    printf("R%dC%d\n",sr,sc);
}
void StaToInt(char *s){
    int sr,sc,len;
    //sr=sc=0;
    len=strlen(s);
    sscanf(s,"R%dC%d",&sr,&sc);
    //printf("%d %d\n",sr,sc);

    int res=sc,mes,k=0;
    //res=sc/26; mes=sc%26;
    char ss[58];
    while(res){
        if(res%26==0){
            ss[k++]='A'+res%26-1+26;
            res/=26;
            res-=1;
        }
        else{
            ss[k++]='A'+res%26-1;
        //printf("%c",'A'+res%26);
            res/=26;
        }
    }
    //ss[k]='A'+res-1;
    for(int i=k-1;i>=0;i--) printf("%c",ss[i]);

    printf("%d\n",sr);
    //printf("R%dC%d\n",sr,sc);
}

int main()
{
    int n;
    char s[100];
    scanf("%d",&n);
    while(n--){
        scanf("%s",s);
        int ok1=0,ok2=0;
        ok1=s[0]=='R'&&isdigit(s[1]);
        for(int i=0;i<strlen(s)-1;i++){
            if(s[i]=='C'&&isdigit(s[i+1])) ok2=1;
        }
        //ok1&ok2判斷選擇哪個函式
        if(ok1&&ok2) StaToInt(s);
        else IntToSta(s);

    }
	return 0;
}