"Recursive and recursive" phone number

"Recursive and recursive" phone number The title describes a number of English letters written below each number on the telephone. The distribution is as follows:
1~abc
2~def
3~ghi
4~ikl
5~mn
6~opq
7~rst
8~uvw
9~xyz
Now given a word list and a string of numbers, please use the words in the word list to translate this password.

Enter the first behavior a positive integer N indicates the number of words in the Word table (n≤100);
Second act a string of numbers not exceeding 100, indicating the password;
The next n rows, each line of a word with a length of not more than 20, represent the Word table.

Output only one line, the translated text, if the password can not be translated, then output "No solutions!", if the password has a variety of translations, the output can be any. Sample input

873373711664thishsthisisbabook

Sample output

Thi Shs b Boo k

Analysis:
Map words to numbers first, then violent match
Code:

#include <iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<algorithm>#include<climits>#include<cstring>#include<string>#include<Set>#include<map>#include<queue>#include<stack>#include<vector>#include<list>#defineRep (I,m,n) for (i=m;i<=n;i++)#defineRSP (It,s) for (Set<int>::iterator It=s.begin (); It!=s.end (); it++)#defineMoD 1000000007#defineINF 0x3f3f3f3f#defineVI vector<int>#definePB Push_back#defineMP Make_pair#defineFi first#defineSe Second#definell Long Long#definePi ACOs (-1.0)Const intmaxn=1e2+Ten;Const intdis[][2]={0,1,-1,0,0,-1,1,0};using namespacestd;ll gcd (ll p,ll q) {returnq==0? P:GCD (q,p%q);} ll Qpow (ll p,ll q) {ll F=1; while(q) {if(q&1) f=f*p%mod;p=p*p%mod;q>>=1;}returnF;}stringA[MAXN],F,B[MAXN];intLen[maxn],len1,n;stringCheckstringAintl) {    stringp;  for(intI=0; i<l;i++)    {        if(a[i]=='a'|| a[i]=='b'|| a[i]=='C') p+='1'; Else if(a[i]=='D'|| a[i]=='e'|| a[i]=='F') p+='2'; Else if(a[i]=='g'|| a[i]=='h'|| a[i]=='I') p+='3'; Else if(a[i]=='J'|| a[i]=='k'|| a[i]=='L') p+='4'; Else if(a[i]=='m'|| a[i]=='N') p+='5'; Else if(a[i]=='P'|| a[i]=='Q'|| a[i]=='o') p+='6'; Else if(a[i]=='R'|| a[i]=='s'|| a[i]=='T') p+='7'; Else if(a[i]=='u'|| a[i]=='v'|| a[i]=='W') p+='8'; Else if(a[i]=='x'|| a[i]=='y'|| a[i]=='Z') p+='9'; }    returnp;}voidDfsintNowstringans) {    if(NOW==LEN1) {COUT&LT;&LT;ANS.SUBSTR (1, Ans.length ()-1) <<endl;exit (0);} Else    {         for(intI=0; i<n;i++)if(Now+len[i]<=len1&&b[i]==f.substr (Now,len[i])) DFS (now+len[i],ans+" "+A[i]); return; }}intMain () {inti,j,m,k,t; scanf ("%d",&N); CIN>>f;len1=f.length (); Rep (I,0, N-1) Cin>>a[i],len[i]=a[i].length (), b[i]=check (a[i],len[i]); DFS (0,""); Puts ("No solutions!"); //System ("pause");    return 0;}

"Recursive and recursive" phone number