sicily 1010. 单词数值

本题主要是Hash思想的应用

课程上机练习题

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1010. 单词数值

Total:

2514

Accepted:

526

	Time Limit: 1sec    Memory Limit:256MB Description 小明喜欢发短信，一天他突然想出一种定义短信中单词的值的方法，也许能够揭示某种规律。 小明的手机键盘如下图所示： 输入法为字母输入，即每次输入一个字母，例如需要输入字母x，由于x在数字键9上，排在第二位，因此需要按下9键两次。 小明定义单词的值如下：单词中每个字母（不区分大小写）的值等于按键数值与按键次数之积，单词的值等于所有字母的值之和。如单词word的值为9+6*3+7*3+3=51。 Input 第一行给出单词数T（1<=T<=100,000）。 接下来T行，每行一个单词，单词最大长度不超过100。 Output 输出T行，每行为单词的对应值。 Sample Input aaarticlea/jpeg;base64,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" alt="" /> Copy sample input to clipboard 3 Word I Day Sample Output 51 12 32 Problem Source: 课程上机练习题

代码：

 // Problem#: 8657
 // Submission#: 2493101
 // The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
 // URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
 // All Copyright reserved by Informatic Lab of Sun Yat-sen University
 #include<iostream>
 #include<map>
 #include<stdio.h>
 #include<string.h>
 using namespace std;
 int main() {
     char alp[] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j',
     'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w','x',
     'y', 'z'};
     int num[] = {, , , ,, , , , , , , , , , ,
     , , , , , , , , , , };
     map<char, int> keyboard;
     for (int i = ; i < ; i++) {
         keyboard[alp[i]] = num[i];
     }
     int words;
     int sum = ;
     char *word;
     word = new char[];
     scanf("%d", &words);
     for (int i = ; i < words; i++) {
         scanf("%s", word);
         int len = strlen(word);
         for (int j = ; j < len; j++) {
             if ( 'A' <= word[j] && word[j] <= 'Z')
                  word[j] = word[j] + ;
             sum += keyboard[word[j]];
         }
         printf("%d\n", sum);
         sum = ;
    }
    return ;
 }