hdu 4287 Intelligent IME map的应用

http://acm.hdu.edu.cn/showproblem.php?pid=4287

题意：

给出n个手机数字序列，m个英文组合，求每个数字序列对应的可能的英文组合在下边的出现了多少个。

思路：

才开始想复杂了，其实只要用一个map即可。

View Code

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <set>

#include <map>

#include <string>



#define CL(a,num) memset((a),(num),sizeof(a))

#define iabs(x)  ((x) > 0 ? (x) : -(x))

#define Min(a,b) (a) > (b)? (b):(a)

#define Max(a,b) (a) > (b)? (a):(b)



#define ll long long

#define inf 0x7f7f7f7f

#define MOD 100000007

#define lc l,m,rt<<1

#define rc m + 1,r,rt<<1|1

#define pi acos(-1.0)

#define test puts("<------------------->")

#define maxn 100007

#define M 100007

#define N 50007

using namespace std;

//freopen("din.txt","r",stdin);



int a[N];

map<int,int>mp;

int n,m;



int find(char *s){

    int i;

    int len = strlen(s);

    int num = 0;

    for (i = 0; i < len; ++i){

        if (s[i] == 'a' || s[i] == 'b' || s[i] == 'c') num = num*10 + 2;

        if (s[i] == 'd' || s[i] == 'e' || s[i] == 'f') num = num*10 + 3;

        if (s[i] == 'g' || s[i] == 'h' || s[i] == 'i') num = num*10 + 4;

        if (s[i] == 'j' || s[i] == 'k' || s[i] == 'l') num = num*10 + 5;

        if (s[i] == 'm' || s[i] == 'n' || s[i] == 'o') num = num*10 + 6;

        if (s[i] == 'p' || s[i] == 'q' || s[i] == 'r' || s[i] == 's') num = num*10 + 7;

        if (s[i] == 't' || s[i] == 'u' || s[i] == 'v') num = num*10 + 8;

        if (s[i] == 'w' || s[i] == 'x' || s[i] == 'y' || s[i] == 'z') num = num*10 + 9;

    }

    return num;

}

int main(){



   //freopen("din.txt","r",stdin);

    int i,t;

    char str[10];

    scanf("%d",&t);

    while (t--){

        scanf("%d%d",&n,&m);

        for (i = 0; i < n; ++i) scanf("%d",&a[i]);

        mp.clear();

        for (i = 0; i < m; ++i){

            scanf("%s",str);

            int x = find(str);

            mp[x]++;

        }

        for (i = 0; i < n; ++i){

            printf("%d\n",mp[a[i]]);

        }

    }

    return 0;

}