USACO 1.3.3 Name That Number 命名那个数字

解析

给定数字，翻译出对应的名字，在有限的名字字典中寻找最终可用正确的名字。
这道题有多种方法，我这里介绍三种：
1. 这是我自己想出来的方法，可是TLE。想法很简单就是，翻译出所有对应的名字，然后依次考察改名字是否在字典中。
我把dict.txt处理成了map，这样在查询的时候就比较方便了。但是显然数字越长，可能情况越多。
2. luogu上看来的，直接把dict.txt处理成数字，然后直接把输入数字来比较，暴力遍历即可，很聪明的解法。
3. 官方题解之一，很精彩，构建字母到数字映射的map[]，对给定数字数组，依次遍历dict.txt中每个name，对每个名字每位翻译再比较，不对直接下一个。

---

代码

1

/* 
PROG:namenum
ID:imking022
LANG:C++
*/ 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

int n;
map<string,int> mapp;
void ini(){
    ifstream fin ("dict.txt");
    string ss;
    while(fin>>ss){
        mapp[ss]++;
    }
}
char reff[10][4] = { "   ","   ","ABC","DEF","GHI",
    "JKL","MNO","PRS",
    "TUV","WXY"
};
int main(void){

    freopen("namenum.out","w",stdout);
    freopen("namenum.in","r",stdin);
    string num,str;
    cin>>num;
    ini();  
    n = num.size();
    int sum = pow(3,n),t,bar,count=0;
    int *cot = (int *) malloc((n+1)*sizeof(int));
    for(int i=0;imemset(cot,0, ((n+1)*sizeof(int))) ;
        t = i;
        for(int j=n;j>=1;j--){
            cot[j] = t%3;
            t-=cot[j];
            t/=3;
        }

        str = "";
        for(int j=1;j<=n;j++){
            bar=num[j-1]-'0';
            str += reff[ bar ][ cot[j] ];   
        }

        if(mapp[str]>0) {
            cout<if(count == 0) cout<<"NONE"<return 0;
}

2

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

int n;
char reff[]="22233344455566670778889990";
string str,s[4700],d[4700];
void ini(){
    ifstream fin ("dict.txt");
    string ss;
    int snum;
    n=0;
    while(fin>>ss){
        s[n] = ss;
        d[n] = ss;
        n++;
    }
    for(int i=0;ifor(int j=0;j'A' ];
    }
}
int main(void){

    freopen("namenum.out","w",stdout);
    freopen("namenum.in","r",stdin);
    cin>>str;
    ini();  

    bool ok = false;
    for(int i=0;iif(s[i] == str) {
            cout<true;
        }
    }

    if(!ok) cout<<"NONE"<return 0;
}

3 这个太优秀了

#include 
#include 
#include 

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;
}