Spreadsheets (26进制转换)

Spreadsheets

Time Limit:10000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

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 follown lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than10⁶ .

Output

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

Sample Input

Input

2R23C55BC23

Output

BC23R23C55题意：定义了两种模式下的字符串下标，要求你转换分析：对于第一种模式，比如R23C55，取出55，那么55实际上是BC的26进制的表示形式，那么只需要按照26进制还原就行，%和/的时候注意一下整除的情况。
打表然后映射也可以

另外要注意的是判断是模式一还是模式二，这里有些坑.

#include<cstring>#include<string>#include<iostream>#include<queue>#include<cstdio>#include<algorithm>#include<map>#include<cstdlib>#include<cmath>#include<vector>//#pragma comment(linker, "/STACK:1024000000,1024000000");using namespace std;#define INF 0x3f3f3f3fchar ans[1000005];char s[1000005];bool judge(){    int len=strlen(s);    int j;    for(j=0;j<len;j++)        if(s[j]>='A'&&s[j]<='Z') continue;        else break;    for(int i=j;i<len;i++)    {        if(s[i]>='0'&&s[i]<='9') continue;        else return true;    }    return false;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%s",s);        int len=strlen(s);        if(judge())        {            int flag=0;            int s1=0,s2=0;            for(int i=1;i<len;i++)            {                if(!flag)                {                    if(s[i]>='0'&&s[i]<='9') s1=s1*10+(s[i]-'0');                    else flag=1;                }                else                {                    if(s[i]>='0'&&s[i]<='9') s2=s2*10+(s[i]-'0');                }            }            int p=0;            while(s2)            {                if(s2%26==0)                {                    ans[p++]='Z';                    s2=(s2-1)/26;                }                else                {                    ans[p++]=(s2%26+'A'-1);                    s2/=26;                }            }            ans[p]='\0';            for(int i=p-1;i>=0;i--)                printf("%c",ans[i]);            printf("%d\n",s1);        }        else        {            int s1=0;            int p;            for(int i=0;i<len;i++)            {                if(s[i]>='0'&&s[i]<='9') s1=s1*10+(s[i]-'0');                if(s[i]>='A'&&s[i]<='Z') p=i;            }            long long temp=0;            for(int i=0;i<=p;i++)            {                temp=temp*26+(s[i]-'A'+1);            }            printf("R%dC%I64d\n",s1,temp);        }    }    return 0;}