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            HDOJ 2714 HDU 2714 ISBN ACM 2714 IN HDU

            Posted on 2010-09-25 20:25 MiYu 閱讀(447) 評(píng)論(1)  編輯 收藏 引用 所屬分類: ACM ( 水題 )

            MiYu原創(chuàng), 轉(zhuǎn)帖請(qǐng)注明 : 轉(zhuǎn)載自 ______________白白の屋    

            題目地址 :

            http://acm.hdu.edu.cn/showproblem.php?pid=2714

            題目描述:

            ISBN

            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
            Total Submission(s): 566    Accepted Submission(s): 172


            Problem Description
            Farmer John's cows enjoy reading books, and FJ has discovered that his cows produce more milk when they read books of a somewhat intellectual nature. He decides to update the barn library to replace all of the cheap romance novels with textbooks on algorithms and mathematics. Unfortunately, a shipment of these new books has fallen in the mud and their ISBN numbers are now hard to read.

            An ISBN (International Standard Book Number) is a ten digit code that uniquely identifies a book. The first nine digits represent the book and the last digit is used to make sure the ISBN is correct. To verify that an ISBN number is correct, you calculate a sum that is 10 times the first digit plus 9 times the second digit plus 8 times the third digit ... all the way until you add 1 times the last digit. If the final number leaves no remainder when divided by 11, the code is a valid ISBN.

            For example 0201103311 is a valid ISBN, since
            10*0 + 9*2 + 8*0 + 7*1 + 6*1 + 5*0 + 4*3 + 3*3 + 2*1 + 1*1 = 55.

            Each of the first nine digits can take a value between 0 and 9. Sometimes it is necessary to make the last digit equal to ten; this is done by writing the last digit as X. For example, 156881111X is a valid ISBN number.

            Your task is to fill in the missing digit from a given ISBN number where the missing digit is represented as '?'.
             

            Input
            * Line 1: A single line with a ten digit ISBN number that contains '?' in a single position
             

            Output
            * Line 1: The missing digit (0..9 or X). Output -1 if there is no acceptable digit for the position marked '?' that gives a valid ISBN.
             

            Sample Input
            15688?111X
             

            Sample Output
            1
             

            Source
             

             

            題目分析 :

            水題, 不錯(cuò)還是 WA 了2次 , 很郁悶 , 主要是沒有注意到 題目描述中點(diǎn)一句話 :

            Each of the first nine digits can take a value between 0 and 9. Sometimes it is necessary to make the last digit equal to ten; this is done by writing the last digit as X. For example, 156881111X is a valid ISBN number.

            這句話是關(guān)鍵 , 意思就是 除了最后一個(gè)數(shù)字 可能是 X 外  其他的都只能為 0-9的數(shù)字.

             

            代碼如下 :

             /*

            Coded By  : MiYu

            Link      : http://www.cnblogs.com/MiYu  || http://m.shnenglu.com/MiYu

            Author By : MiYu

            Test      : 1

            Program   : 2714

            */

            //#pragma warning( disable:4789 )

            #include <iostream>

            #include <algorithm>

            #include <string>

            #include <set>

            #include <map>

            #include <utility>

            #include <queue>

            #include <stack>

            #include <list>

            #include <vector>

            #include <cstdio>

            #include <cstdlib>

            #include <cstring>

            #include <cmath>

            using namespace std;

            int main ()

            {

                string str;

                while ( cin >> str )

                {

                      int N = str.size();

                      int sum = 0, ask = 0;

                      for ( int i = 1; i <= N; ++ i )

                      {

                           switch ( str[i-1] )

                           {

                                   case 'X':

                                            sum += (10-i+1) * 10;  break;

                                   case '?':

                                            ask = 10-i+1;  break;

                                   default :

                                            sum += (10-i+1) * (str[i-1] - '0');  break;      

                           }

                      } 

                      int pos = -1;

                      for ( int i = 0; i <= 10; ++ i )  

                      {

                           if ( ( sum + ask * i ) % 11 == 0 )

                           {

                                pos = i;

                                break;     

                           }    

                      } 

                      if ( ask != 1 && pos == 10 ) pos = -1; //沒有這句代碼 WA  2次  很郁悶.

                      if ( pos == 10 )

                         cout << 'X' << endl;  

                      else

                          cout << pos << endl;

                }

                return 0;

            }


             

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            2012-03-09 21:40 by ACM小鳥~
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