In the present world you frequently meet a lot of call numbers and they are going to be longer and longer. You need to remember such a kind of numbers. One method to do it in an easy way is to assign letters to digits as shown in the following picture:
1 ij 2 abc 3 def
4 gh 5 kl 6 mn
7 prs 8 tuv 9 wxy
0 oqz
This way every word or a group of words can be assigned a unique number, so you can remember words instead of call numbers. It is evident that it has its own charm if it is possible to find some simple relationship between the word and the person itself. So you can learn that the call number 941837296 of a chess-playing friend of yours can be read as WHITEPAWN, and the call number 2855304 of your favourite teacher is read BULLDOG.
Write a program to find the length of shortest sequence of words (i.e. one having the smallest possible number of words) which corresponds to a given number and a given list of words. The correspondence is described by the picture above.
Input
Input contains a series of tests. The first line of each test contains the call number, the transcription of which you have to find. The number consists of at most 100 digits.
The second line contains the total number of the words in the dictionary (maximum is 50 000).
Each of the remaining lines contains one word, which consists of maximally 50 small letters of the English alphabet.
The total size of the input doesn’t exceed 300 KB. The last line contains call number −1.
Output
Print the length of shortest sequence of words, which means if the shortest sequence consists 3 words, output 3.
If there is no solution to the input data, the line contains text “No solution.”.
答案：
#include <iostream>
#include <string.h>
#include <vector>
using namespace std;
vector<string> outstr;
int Checkstr(){
string phonenum ;
cin>>phonenum;
if(phonenum.size() == 2 and phonenum[0] == '-' and phonenum[1] == '1'){
return -1;
}
int count;
cin>>count;
vector <string> instr;
vector <int> numstr;
int sequence = 0;
for(int i = 0 ; i < count ; i++){
string tstr;
cin>>tstr;
instr.insert(instr.end(), tstr);
}
for(int i = 0 ; i < instr.size() ; i++){
int tse = 0;
for(int j = 0 ; j < instr[i].size() ; j++){
int instrsq = -1;
switch (instr[i][j]) {
case 'i':;
case 'j':
instrsq = 1;
break;
case 'a':;
case 'b':;
case 'c':
instrsq = 2;
break;
case 'd':;
case 'e':;
case 'f':
instrsq = 3;
break;
case 'g':;
case 'h':
instrsq = 4;
break;
case 'k':;
case 'l':
instrsq = 5;
break;
case 'm':;
case 'n':
instrsq = 6;
break;
case 'p':;
case 'r':;
case 's':
instrsq = 7;
break;
case 't':;
case 'u':;
case 'v':
instrsq = 8;
break;
case 'w':;
case 'x':;
case 'y':
instrsq = 9;
break;
case 'o':;
case 'q':;
case 'z':
instrsq = 0;
break;
default:
break;
}
if(instrsq >=0 and (phonenum[sequence+j]-'0') == instrsq and tse != 1){
tse = 0;
cout<<"instr1 is "<<instr[i][j]<<endl;
}else{
cout<<"instr2 is "<<instr[i][j]<<endl;
tse = 1;
break;
}
}
if(tse == 0){
sequence = (int)instr[i].size() + sequence;
outstr.insert(outstr.end(), instr[i]);
}
}
return 0;
}
int main(int argc, const char * argv[]) {
vector<string> ostr;
while (1) {
if(Checkstr() == -1){
for(int i = 0 ; i < ostr.size() ; i++){
cout<<ostr[i]<<endl;
}
break;
}
else{
string stemp;
if(outstr.size()<1){
ostr.insert(ostr.end(), "No solution.");
}
else{
for(int i = 0 ; i < outstr.size() ; i++){
stemp = stemp + outstr[i] + ' ';
}
ostr.insert(ostr.end(), stemp);
outstr.clear();
}
}
}
return 0;
}