ZCMU-1770-打字

1770: 打字

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 80   Solved: 15
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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

【解析】

这道题的意思就是首先输入n组数组，然后输入一下我们需要进行统计这个操作能打印出多少字符的m,再输入有k个字

符。就是问在这k个字符当中我每次的操作能打印出几个。其实我们只要统计下字符能通过哪些操作打印出来就好了比

如go就可以通过46打印出来，in也是46，gn也是46所以46操作能打印出来的是3个字符。我们可以建一个数组把每个字

符对应的操作数存储起来。然后最后我们就可以直接进行查询，这个操作数有没有值。此处是模仿他人代码

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
#include <map>
using namespace std;
int a[28]={2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,9,9,9,9};//每个字符对应的操作
int b[100000];
int c[6000];
char s[100000];
int main()
{
	int t,i,j;
	scanf("%d",&t);
	int n,m;
    while(t--)
    {
        scanf("%d%d",&n,&m);
		memset(b,0,sizeof(b));
		memset(c,0,sizeof(c));
		for(i=0;i<n;i++)
        {
            scanf("%d",&c[i]);//输入操作数
        }
		for(i=0;i<m;i++)
        {
        	scanf("%s",s);
			int sum=0;
            for(j=0;j<strlen(s);j++)
				sum=sum*10+a[s[j]-97];//计算每个字符的操作数
			b[sum]++;
        }
        for(i=0;i<n;i++)
        {
        printf("%d\n",b[c[i]]);//在操作数中输出值
        }
    }
    return 0;
}