cf # 1 B. Spreadsheets

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分类专栏： *******数学******** ------水题 ~~~~程序设计~~~~

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本文介绍了如何将C++代码中表示单元格坐标的方式从RC形式转换为字母+数字形式，并详细解释了两种数制之间的转换逻辑与实现过程。

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

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.

Sample test(s)

input

2 R23C55 BC23

output

BC23 R23C55
对一个网格有两种表示形式，一个是R***C***形式，另一个是字母 +数字形式，两个转换关系可以看样例；；

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int myabs(int a,int b)
{
    if(a>b)
        return a-b;
    else
        return b-a;
}
int f(char s[])
{
    int ans=0;
    for(int i=0; i<strlen(s); i++)
    {
        ans=(s[i]-'0')+ans*10;
    }
    return ans;
}
int altonum(char s[])
{
    int ans=0;
    for(int i=0; i<strlen(s); i++)
    {
        ans=ans*26+(int)(s[i]-'A'+1);
    }

    return ans;

}
int main()
{
    char ans1[50],ans2[50];
    char str[100];
    int ci,ri;
    int n,ans;
    int k,g;
    int num[100];
    cin>>n;
    while(n--)
    {
        cin>>str;
        int len=strlen(str);
        ci=-1;
        ri=-1;
        for(int i=0; i<len; i++)
        {
            if(str[i]=='R')
                ri=i;
            if(str[i]=='C')
                ci=i;
        }
        if(myabs(ci,ri)>1&&str[ci-1]-'0'>=0&&str[ci-1]-'0'<=9&&ci!=-1&&ri!=-1)///R C 之间必须是数字，
        {
            //cout<<1<<endl;
            k=0,g=0;
            int j;
            for(int i=0; i<len; i++)
            {
                if(str[i]=='R')
                {
                    for(j=i+1; j<len; j++)
                    {
                        if(str[j]=='C')
                        {
                            //  cout<<"j"<<j<<endl;
                            ans2[k]='\0';
                            break;
                        }
                        ans2[k++]=str[j];
                    }
                    for(j=j+1; j<len; j++)
                    {
                        ans1[g++]=str[j];
                    }
                    ans1[g]='\0';
                    break;
                }

            }
            ans=f(ans1);
            k=0;
            while(ans)
            {
                num[k++]=ans%26;
                int t=ans;
                ans=ans/26;
                if(t%26==0)
            ans--;
            }
            for(int i=k-1; i>=0; i--)
            {
                if(num[i]==0)
                    cout<<"Z";
                else
                cout<<char(num[i]-1+'A');
            }
            cout<<ans2<<endl;
        }
        else
        {
            k=0,g=0;
            for(int i=0; i<len; i++)
            {
                if(isalpha(str[i]))
                    ans2[k++]=str[i];
                else if(str[i]-'0'<=9&&str[i]-'0'>=0)
                    ans1[g++]=str[i];
            }
            ans1[g]='\0';
            ans2[k]='\0';
            cout<<"R"<<ans1<<"C";
            ans=altonum(ans2);
            cout<<ans<<endl;
        }

    }
    return 0;
}