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?

　　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.

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

1
3 5
46
64448
74
go
in
night
might
gn

3
2
0

2012 ACM/ICPC Asia Regional Tianjin Online

liuyiding

 

题意：

　　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

 

a 是2 b是2 c 是2  d是3 e是3.。。。。。。。

把字符串转化以后的数字串  看在上面出现了几次

理所当然 map解决

这题主要是练习 map 和char[] 的合作

 

#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<map>
using namespace std;
char key[27]="22233344455566677778889999";
char num[5005][10],word[5005][10];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m,i,j,len;
        map <string,int> M;
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
        {
            scanf("%s",num[i]);
        }
        map<string,int>::iterator it; 
        for(i=0;i<m;i++)
        {
            scanf("%s",word[i]);
            len=strlen(word[i]);
            for(j=0;j<len;j++)
                word[i][j]=key[word[i][j]-'a'];
            it=M.find(word[i]);
            if(it==M.end())
                M[word[i]]=1;
            else
                M[word[i]]++;
        }    
        for(i=0;i<n;i++)
        {
            it=M.find(num[i]);
            if(it==M.end())
                printf("0\n");
            else
                printf("%d\n",M[num[i]]);
        }
    }
    return 0;
}