USACO Name That Number

1、这道题刚开始看到要从文件读入“字典”，有点心虚，其实比较简单。

2、思路就是暴力枚举~刚开始我想枚举数字或字母，比较麻烦，后来借鉴了USACO上面讲解的某道IOI题目的思路，换了枚举对象，变成枚举单词（因为题目说了不到5000个），程序一下变得很容易。

/*
ID:mrxy564
PROG:namenum
LANG:C++
*/
#include<cstdio>
#include<map>
#include<cstring>
using namespace std;
map<char,int> a;
int main()
{
    char dic[5000][15];
    a['A']=2;a['B']=2;a['C']=2;
    a['D']=3;a['E']=3;a['F']=3;
    a['G']=4;a['H']=4;a['I']=4;
    a['J']=5;a['K']=5;a['L']=5;
    a['M']=6;a['N']=6;a['O']=6;
    a['P']=7;a['R']=7;a['S']=7;
    a['T']=8;a['U']=8;a['V']=8;
    a['W']=9;a['X']=9;a['Y']=9;
    freopen("namenum.in","r",stdin);
    freopen("namenum.out","w",stdout);
    FILE *fin;
    fin=fopen("dict.txt","rb");
    int i=0,cnt;
    while(fscanf(fin,"%s",dic[i])==1) i++;
    cnt=i;
    fclose(fin);
    long long num;
    char s[15];
    while(scanf("%lld",&num)==1){
         sprintf(s,"%lld",num);
         int len=strlen(s);
         bool flag,judge=false;
         for(int i=0;i<cnt;i++){
            flag=true;
         for(int j=0;j<len;j++)
             if((s[j]-'0')!=a[dic[i][j]]){
                 flag=false;
                 break;
             }
             if(flag&&dic[i][len]=='\0'){
               printf("%s\n",dic[i]);
               judge=true;
             }
          }
          if(!judge) printf("NONE\n");
    }
    return 0;
}
官方题解：

There are two ways to do this problem. One is, given the number, to generate all possible strings that encode to that number and look them up in the dictionary. Since there are 3 letters for each number and 12 digits in the string, that's 3¹² = 531441 lookups into a dictionary of size 5000, which although manageable would be a little on the long side (binary search can help this).

Or, we can examine each word in the dictionary to see if it maps to the digits of the number in question. This has the the advantage of a shorter program that probably will work right first time.

---

 

Here is Argentina competitor's Michel Mizrah's solution using the first method with a binary search. While it is blazingly fast, it does have the disadvantage of some fairly tricky coding in the binary search routine. A single off-by-one error would doom a program in a contest.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char num[12],sol[12];
char dict[5000][13];
int nsolutions = 0;
int nwords;
int maxlen;
FILE *out;

void calc (int charloc, int low, int high) {
    if (charloc == maxlen) {
        sol[charloc] = '\0';
        for (int x = low; x < high; x++) {
            if (strcmp (sol, dict[x]) == 0) {
                fprintf (out, "%s\n", sol);
                nsolutions++;
            }
        }
        return;
   }
   if (charloc > 0) {
        for (int j=low; j <= high; j++){
            if (sol[charloc-1] == dict[j][charloc-1]) {
                low=j;
                while (sol[charloc-1] == dict[j][charloc-1])
                    j++;
                high=j;
                break;
            }
            if (j == high) return;
        }
    }
    if (low > high) return;
    switch(num[charloc]){
      case '2':sol[charloc] = 'A'; calc(charloc+1,low,high);
               sol[charloc] = 'B'; calc(charloc+1,low,high);
               sol[charloc] = 'C'; calc(charloc+1,low,high);
               break; 
      case '3':sol[charloc] = 'D'; calc(charloc+1,low,high);
               sol[charloc] = 'E'; calc(charloc+1,low,high);
               sol[charloc] = 'F'; calc(charloc+1,low,high);
               break; 
      case '4':sol[charloc] = 'G'; calc(charloc+1,low,high);
               sol[charloc] = 'H'; calc(charloc+1,low,high);
               sol[charloc] = 'I'; calc(charloc+1,low,high);
               break; 
      case '5':sol[charloc] = 'J'; calc(charloc+1,low,high);
               sol[charloc] = 'K'; calc(charloc+1,low,high);
               sol[charloc] = 'L'; calc(charloc+1,low,high);
               break; 
      case '6':sol[charloc] = 'M'; calc(charloc+1,low,high);
               sol[charloc] = 'N'; calc(charloc+1,low,high);
               sol[charloc] = 'O'; calc(charloc+1,low,high);
               break; 
      case '7':sol[charloc] = 'P'; calc(charloc+1,low,high);
               sol[charloc] = 'R'; calc(charloc+1,low,high);
               sol[charloc] = 'S'; calc(charloc+1,low,high);
               break; 
      case '8':sol[charloc] = 'T'; calc(charloc+1,low,high);
               sol[charloc] = 'U'; calc(charloc+1,low,high);
               sol[charloc] = 'V'; calc(charloc+1,low,high);
               break; 
      case '9':sol[charloc] = 'W'; calc(charloc+1,low,high);
               sol[charloc] = 'X'; calc(charloc+1,low,high);
               sol[charloc] = 'Y'; calc(charloc+1,low,high);
               break;
   }
}

int main(){
    FILE *in=fopen ("namenum.in", "r");
    FILE *in2=fopen ("dict.txt", "r");
    int j;
    out=fopen ("namenum.out","w");
    for (nwords = 0; fscanf (in2, "%s", &dict[nwords++]) != EOF; )
        ;
    fscanf (in, "%s",&num);
    maxlen = strlen(num);
    calc (0, 0, nwords);
    if (nsolutions == 0) fprintf(out,"NONE\n");
    return 0;
}

---

 

The solution below might be considered to be a bit more straightforward: no tricky offsets, no +1 or -1, no knowledge about character values. The lines of actual code in this solution are minimal.

This is the sort of program that might work reliably the first time and every time. The only tricky part is knowing that scanf will yield string without a newline on the end:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main() {
    FILE *in = fopen ("namenum.in", "r");
    FILE *in2 = fopen ("dict.txt", "r");
    FILE *out = fopen ("namenum.out","w");
    int nsolutions = 0;
    int numlen;
    char word[80], num[80], *p, *q, map[256];
    int i, j;
    map['A'] = map['B'] = map['C'] = '2';
    map['D'] = map['E'] = map['F'] = '3';
    map['G'] = map['H'] = map['I'] = '4';
    map['J'] = map['K'] = map['L'] = '5';
    map['M'] = map['N'] = map['O'] = '6';
    map['P'] = map['R'] = map['S'] = '7';
    map['T'] = map['U'] = map['V'] = '8';
    map['W'] = map['X'] = map['Y'] = '9';
    fscanf (in, "%s",num);
    numlen = strlen(num);
    while (fscanf (in2, "%s", word) != EOF) {
        for (p=word, q=num; *p && *q; p++, q++) {
            if (map[*p] != *q)
                break;
        }
        if (*p == '\0' && *q == '\0') {
            fprintf (out, "%s\n", word);
            nsolutions++;
        }
    }
    if (nsolutions == 0) fprintf(out,"NONE\n");
    return 0;
}