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            posts - 21, comments - 2, trackbacks - 0, articles - 0

            POJ 2273 An Excel-lent Problem

            Posted on 2011-09-05 21:53 acpeng 閱讀(487) 評(píng)論(0)  編輯 收藏 引用 所屬分類: ACM程序

            把整數(shù)轉(zhuǎn)化為excel對(duì)應(yīng)的列值http://poj.org/problem?id=2273
            Description

            A certain spreadsheet program labels the columns of a spreadsheet using letters. Column 1 is labeled as ``A", column 2 as ``B", ..., column 26 as ``Z". When the number of columns is greater than 26, another letter is used. For example, column 27 is ``AA", column 28 is ``AB" and column 52 is ``AZ". It follows that column 53 would be ``BA" and so on. Similarly, when column ``ZZ" is reached, the next column would be ``AAA", then ``AAB" and so on.

            The rows in the spreadsheet are labeled using the row number. Rows start at 1.

            The designation for a particular cell within the spreadsheet is created by combining the column label with the row label. For example, the upper-left most cell would be ``A1". The cell at column 55 row 23 would be ``BC23".

            You will write a program that converts numeric row and column values into the spreadsheet designation.

            Input

            Input consists of lines of the form: RnCm. n represents the row number [1,300000000] and m represents the column number, 1<=m<=300000000. The values n and m define a single cell on the spreadsheet. Input terminates with the line: R0C0 (that is, n and m are 0). There will be no leading zeroes or extra spaces in the input.

            Output

            For each line of input (except the terminating line), you will print out the spreadsheet designation for the specified cell as described above.

            Sample Input

            R1C1
            R3C1
            R1C3
            R299999999C26
            R52C52
            R53C17576
            R53C17602
            R0C0
            

            Sample Output

            A1
            A3
            C1
            Z299999999
            AZ52
            YYZ53
            YZZ53
            

            Source

            Greater New York 2004

            進(jìn)制轉(zhuǎn)化問題,把列數(shù)改成26進(jìn)制輸出,分別用A,B...Z代替1,2,3...這里base[0]='Z',表示n與26余數(shù)為0時(shí)的取值,但進(jìn)位問題需要注意,分余數(shù)是否為0兩種情況分析。
            代碼:
            #include<stdio.h>
            #define R 26
            char base[30]="ZABCDEFGHIJKLMNOPQRSTUVWXYZ";
            void fun(int N)
            {
                
            int i=0,j; char str[10];
                
            while(N>R)
                
            {
                    str[i
            ++]=base[N%R];
                    N
            =N%R==0?N/R-1:N/R;
                }

                str[i]
            =base[N];
                
            for(j=i;j>=0;j--)
                    printf(
            "%c",str[j]);
            }

            int main()
            {
                
            int m,n;
                
            while(scanf("%*c%d%*c%d",&m,&n)!=EOF)
                
            {
                    getchar();
                    
            if(m==0 && n==0)break;
                    fun(n);
                    printf(
            "%d\n",m);
                }

                
            return 0;
            }

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