codeforces 1B 解题报告

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来自Vic_:http://blog.csdn.net/Vic___

每周一道算法题安慰自己竞赛的心，囧

1B一个月前做的，当时专牛角尖，一个月后，昨天看组成原理，讲到进制转换的时候想起这道题了，突然灵光一闪，做出来了

原题地址：http://codeforces.com/problemset/problem/1/B

问题描述

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
2R23C55BC23
output
BC23R23C55

题目理解

学点英语

spreadsheets n. 电子表格；电子数据表；试算表；空白表格程序

numeration n. 计算；数字的读法；[数] 命数法；编号

sequence n. [数][计] 序列；顺序；续发事件
coordinates n. [数] 坐标；相配之衣物

我的理解

这题是一个格式转换题目

在电子表格中，坐标一行是用字母来表示，一列是用数字表示的，这样好表示每一个单元格

那上图中的A1 另一种表示就是R1C1

R ：row

C ：column

说白了就是字母转换成数字或者数字转换成字母

字母转换数字

通过上图分析，显而易见，字母转数字，第n字母就是26的n-1权重，通过权重展开即可求得数字

数字转字母

上图可见，这其实26进制的转换，所以可以反复除26求余得到字母。

但是请注意，52转换，52/26商2……余0，结果为B0，但是字母没有0的表示，所以，当发生余数为0时要处理

余数为0，我的处理方法是：

当余数为0，自动转换成Z，商减一

验证：702/26 商27……余0，0转换为Z，27-1=26；结果为ZZ。成立

解题代码

#include <stdio.h>#include <stdlib.h>const char Change[] ="ZABCDEFGHIJKLMNOPQRSTUVWXY";//字母 与 数字 转换void main (){    int n;    scanf("%d",&n);    while(n--)    {        char a[100] ;        scanf("%s",a);        int flag=0;        int locc=2;        if(a[0]=='R'&&a[1]>='0'&&a[1]<='9')        {            while(a[locc]!='\0')//find C            {                if(a[locc]!='C'){                locc++;                }                else{                    flag=1;                    break;                }            }        }        if(flag)        {            char b[100],c[100];            int b_l=0,c_l=0;            int i;            for(i =1 ; i < locc;i++)            {                c[c_l++]=a[i];            }            c[c_l]='\0';            for(i = locc + 1 ; a[i]!='\0' ;i++)            {                b[b_l++]=a[i];            }            b[b_l]='\0';            long temp = atoi(b);            b[0] = '\0';            b_l=0;            while(temp>26)            {                if(temp%26==0)                {                    b[b_l++]=Change[0]; //特殊处理                    temp=temp/26 - 1;                }                else{                b[b_l++]=Change[temp%26];                temp=temp/26;                }            }            b[b_l++]=Change[temp%26];            b[b_l]='\0';            for(i = 0 ; i < b_l / 2 ; i ++)            {                char te;                te= b[i];                b[i]=b[b_l-i-1];                b[b_l-i-1]=te;            }            printf("%s%s\n",b,c);        }        else        {            char b[100],c[100];            int b_l=0,c_l=0;            int i ;            for(i = 0 ; a[i] !='\0';i++)            {                if(a[i]<'0'||a[i]>'9')                {                    b[b_l++] =a[i];                }                else                {                    c[c_l++] =a[i];                }            }            b[b_l]='\0';            c[c_l]='\0';            long result= 0;            long temp = 1 ;            for (i = b_l-1;i >= 0; i --)            {                result = result + ((b[i] - 'A') + 1)*temp;                temp = temp *26;            }            printf("R%sC%d\n",c,result);        }    }}