USACO 1.2 Name That Number （AD-hoc)

一个字母只对应一个数字，从字典中读入一个单词，把它转化成唯一对应的数字，看它是否与给出的数字匹配，时间规模是5000*12=6e4，空间规模是常数，而且编程复杂度较低.

一开始，我把对应的数字设为int，后来发现，溢出了，囧。改成string字符数组就好了

/*
  ID:twd30651
  PROG:namenum
  LANG:C++
*/
#include<iostream>
#include<fstream>
#include<string.h>
#include<string>
using namespace std;
//          2: A,B,C     5: J,K,L    8: T,U,V
//          3: D,E,F     6: M,N,O    9: W,X,Y
//          4: G,H,I     7: P,R,S
//       a b c d e f g h i j k l m n o p q r s t u v w x y
int m[]={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};
typedef struct node
{
    char name[20];
    string num;
}node;
node names[5000];
int nl;
void generate()
{
    for(int i=0;i<nl;++i)
    {
        names[i].num="";
        for(unsigned j=0;j<strlen(names[i].name);++j)
        {
            if(names[i].name[j]=='Q')continue;
            names[i].num+=m[names[i].name[j]-'A']+'0';
        }
    }
}
int main(int argc,char *argv[])
{
    freopen("namenum.in","r",stdin);
     freopen("namenum.out","w",stdout);
    FILE *f=fopen("dict.txt","r");
    int i=0;
    while(fscanf(f,"%s",names[i].name)==1)
    {
        if(names[i].name[0]=='Z')break;
        i++;
    }
    string NUM;
    cin>>NUM;
    nl=i;
    generate();
    int flag=0;
    for(int i=0;i<nl;++i)
    {
        if(names[i].num==NUM){
            flag=1;
            printf("%s\n",names[i].name);
        }
    }
    if(flag==0)
        printf("NONE\n");
    return 0;
}