"Recursive and recursive" phone number

Question L: "Recursive and recursive" phone number Title DescriptionA number of English letters are 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.

InputThe 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.

Outputonly 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

#include <bits/stdc++.h>using namespacestd;intn,cnt =1;structWord {strings,l;} a[101];stack<int>STA;stringO;stack<int>p;voidSlove (intYintx) {if(!cnt)return ; if(Y = =o.size ()) {         while(!Q.empty ()) {            intx =P.top ();            Sta.push (x);        P.pop (); }         intx =Sta.top ();            Sta.pop (); cout<<A[x].s;  while(!Sta.empty ()) {            intx =Sta.top ();            Sta.pop (); cout<<" "<<A[x].s; } cout<<Endl; CNT=0; return ; }    intFlag;  for(inti = x; I < n; i++) {flag=1;  for(intj =0; J < A[i].l.size (); J + +) {            if(A[i].l[j]! = o[y+J]) {flag=0;  Break; }        }        if(flag) {P.push (i);//cout<< "+" <<a[i].s<<endl;Slove (Y+a[i].l.size (),0) ; }    }     if(!flag&&CNT) {        if(P.empty ()) {printf ("No solutions!\n"); return ; } Else {            intx =p.top ();//cout<< "-" <<a[x].s<<endl;P.pop (); Slove (y-a[x].l.size (), x+1); }    }}intMain () {//freopen ("Data.in.txt", "w", stdout);Cin>>N; CIN>>o;  while(!p.empty ()) P.pop ();  for(inti =0; I < n; i++) {cin>>A[i].s;  for(intj =0; J < A[i].s.size (); J + +) {            intm; if(A[i].s[j] >='o') M= (A[i].s[j] +1-'a')/3+1; Elsem = (A[i].s[j]-'a')/3+1; A[I].L+ = (m +'0'); }} slove (0,0); return 0;}

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"Recursive and recursive" phone number