J - Intelligent IME---（2015 summer training #3）

J - Intelligent IME

时限:1000MS     内存:32768KB     64位IO格式:%I64d & %I64u

问题描述

　　We all use cell phone today. And we must be familiar with the intelligent English input method on the cell phone. To be specific, the number buttons may correspond to some English letters respectively, as shown below: 
　　2 : a, b, c    3 : d, e, f    4 : g, h, i    5 : j, k, l    6 : m, n, o     
　　7 : p, q, r, s  8 : t, u, v    9 : w, x, y, z 
　　When we want to input the word “wing”, we press the button 9, 4, 6, 4, then the input method will choose from an embedded dictionary, all words matching the input number sequence, such as “wing”, “whoi”, “zhog”. Here comes our question, given a dictionary, how many words in it match some input number sequences? 

 

输入

　　First is an integer T, indicating the number of test cases. Then T block follows, each of which is formatted like this: 
　　Two integer N (1 <= N <= 5000), M (1 <= M <= 5000), indicating the number of input number sequences and the number of words in the dictionary, respectively. Then comes N lines, each line contains a number sequence, consisting of no more than 6 digits. Then comes M lines, each line contains a letter string, consisting of no more than 6 lower letters. It is guaranteed that there are neither duplicated number sequences nor duplicated words. 

 

输出

　　For each input block, output N integers, indicating how many words in the dictionary match the corresponding number sequence, each integer per line. 

 

样例输入

      
       1
3 5
46
64448
74
go
in
night
might
gn 
     
     

 

样例输出

      
       3
2
0 
     
     

分析：手机键盘，给你一个数，然后给你很多字符串，问你用这个数敲打键盘，能打出多少给定的字符串？

思路：根据每个给出的字符串，写出相应的数字，然后和输入的比较是否相等。（哈希算法）

CODE：

#include <iostream>
#include <string.h>
using namespace std;

int ans[1000005];

int main()
{
    int key[127];
    key['a']=key['b']=key['c']=2;
    key['d']=key['e']=key['f']=3;
    key['g']=key['h']=key['i']=4;
    key['j']=key['k']=key['l']=5;
    key['m']=key['n']=key['o']=6;
    key['p']=key['q']=key['r']=key['s']=7;
    key['t']=key['u']=key['v']=8;
    key['w']=key['x']=key['y']=key['z']=9;
    int t;
    cin>>t;
    while(t--){
        int n,m,a[5005];
        memset(ans,0,sizeof(ans));
        char b[5005][8];
        cin>>n>>m;
        for(int i=0;i<n;i++) cin>>a[i];
        for(int i=0;i<m;i++){
            cin>>b[i];
            int sum=0;
            for(int j=0;j<strlen(b[i]);j++)
                    sum=sum*10+key[(int)b[i][j]];
            ans[sum]++;
        }
        for(int i=0;i<n;i++)
            cout<<ans[a[i]]<<endl;
    }
    return 0;
}