hdu 4287Intelligent IME(简单hash)

Intelligent IME

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1810    Accepted Submission(s): 897

Problem Description

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?

 

Input

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.

 

Output

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

 

Sample Input

1
3 5
46
64448
74
go
in
night
might
gn

 

Sample Output

3
2
0

 

Source

2012 ACM/ICPC Asia Regional Tianjin Online

 

 
题目意思就不多说了，将下面的字符串hash映射到上面的数字上即可。

AC代码：

#include<iostream>
#include<cstring>
using namespace std;
int hash1[1000005];
int a[5005];
char b[5005][8];

int cal(char *s)
{
    int tmp=0;
    int len=strlen(s),i;
    for(i=0;i<len;i++)
    {
        if(s[i]>='a'&&s[i]<='c')       tmp=tmp*10+2;
        else if(s[i]>='d'&&s[i]<='f') tmp=tmp*10+3;
        else if(s[i]>='g'&&s[i]<='i') tmp=tmp*10+4;
        else if(s[i]>='j'&&s[i]<='l') tmp=tmp*10+5;
        else if(s[i]>='m'&&s[i]<='o') tmp=tmp*10+6;
        else if(s[i]>='p'&&s[i]<='s') tmp=tmp*10+7;
        else if(s[i]>='t'&&s[i]<='v') tmp=tmp*10+8;
        else if(s[i]>='w'&&s[i]<='z') tmp=tmp*10+9;
    }
    return tmp;
}

int main()
{
    int tes,i;
    cin>>tes;
    while(tes--)
    {
        int n,m;
        cin>>n>>m;
        for(i=0;i<n;i++)
            cin>>a[i];   //存放数字
        for(i=0;i<m;i++)
            cin>>b[i];
        memset(hash1,0,sizeof(hash1));
        for(i=0;i<m;i++)
            hash1[cal(b[i])]++;
        for(i=0;i<n;i++)
            cout<<hash1[a[i]]<<endl;
    }
    return 0;
}

//406MS