Pat class B 1109 good at C (detailed)

When you are asked by the interviewer to use C Write a “Hello World” when , Do you have the ability to write one out as shown in the figure below ？

Input format ：

The input first gives 26 English capital letters A-Z, One for each letter  7×5  Of 、 from  C  and  .  The matrix formed by . Finally, give a sentence in one line , End with a carriage return . A sentence consists of several words （ Each contains no more than 10 Consecutive capital letters ） Composed of , Words are separated by any non capitalized English letter .

The title is guaranteed to give at least one word .

Output format ：

For each word , Output each letter in a row in matrix form , There is a column of spaces between the letters . There must be no extra spaces at the beginning and end of the word .

There must be a space between two adjacent words . There must be no extra blank lines at the beginning and end of the output .

sample input ：

..C..
.C.C.
C...C
CCCCC
C...C
C...C
C...C
CCCC.
C...C
C...C
CCCC.
C...C
C...C
CCCC.
.CCC.
C...C
C....
C....
C....
C...C
.CCC.
CCCC.
C...C
C...C
C...C
C...C
C...C
CCCC.
CCCCC
C....
C....
CCCC.
C....
C....
CCCCC
CCCCC
C....
C....
CCCC.
C....
C....
C....
CCCC.
C...C
C....
C.CCC
C...C
C...C
CCCC.
C...C
C...C
C...C
CCCCC
C...C
C...C
C...C
CCCCC
..C..
..C..
..C..
..C..
..C..
CCCCC
CCCCC
....C
....C
....C
....C
C...C
.CCC.
C...C
C..C.
C.C..
CC...
C.C..
C..C.
C...C
C....
C....
C....
C....
C....
C....
CCCCC
C...C
C...C
CC.CC
C.C.C
C...C
C...C
C...C
C...C
C...C
CC..C
C.C.C
C..CC
C...C
C...C
.CCC.
C...C
C...C
C...C
C...C
C...C
.CCC.
CCCC.
C...C
C...C
CCCC.
C....
C....
C....
.CCC.
C...C
C...C
C...C
C.C.C
C..CC
.CCC.
CCCC.
C...C
CCCC.
CC...
C.C..
C..C.
C...C
.CCC.
C...C
C....
.CCC.
....C
C...C
.CCC.
CCCCC
..C..
..C..
..C..
..C..
..C..
..C..
C...C
C...C
C...C
C...C
C...C
C...C
.CCC.
C...C
C...C
C...C
C...C
C...C
.C.C.
..C..
C...C
C...C
C...C
C.C.C
CC.CC
C...C
C...C
C...C
C...C
.C.C.
..C..
.C.C.
C...C
C...C
C...C
C...C
.C.C.
..C..
..C..
..C..
..C..
CCCCC
....C
...C.
..C..
.C...
C....
CCCCC
HELLO~WORLD!

sample output ：

C...C CCCCC C.... C.... .CCC.
C...C C.... C.... C.... C...C
C...C C.... C.... C.... C...C
CCCCC CCCC. C.... C.... C...C
C...C C.... C.... C.... C...C
C...C C.... C.... C.... C...C
C...C CCCCC CCCCC CCCCC .CCC.

C...C .CCC. CCCC. C.... CCCC.
C...C C...C C...C C.... C...C
C...C C...C CCCC. C.... C...C
C.C.C C...C CC... C.... C...C
CC.CC C...C C.C.. C.... C...C
C...C C...C C..C. C.... C...C
C...C .CCC. C...C CCCCC CCCC.

 

This is a slightly complicated string processing problem , Take more time to get familiar with this practice .

AC Code ：

#include<bits/stdc++.h>
using namespace std;

string alpha[66][66];  // Read in 26 Letters 
string ans[6666];    // Used to output results , Pay attention to the space 
string targetStr;      // Target string 

int main(){
	for(int i = 0; i < 26; i++) //26 Letters 
		for(int j = 0; j < 7; j++)	// Each letter has 7 That's ok , So I have to save 7 That's ok 
			getline(cin, alpha[i][j]); // Save line by line 
	getline(cin, targetStr); // Save the target string 
	
	int idx = 0; // The subscript of the result array 
	int t = 0;	// Used to wrap lines 
	bool f = false; // Determine whether the next digit is a letter 
	for(int i = 0; i < targetStr.length(); i++){ 
		if(targetStr[i] >= 'A' and targetStr[i] <= 'Z'){ // If it's a letter 
			for(int j = 0; j < 7; j++){ // Read a letter , Because a letter has 7 That's ok , So read it 7 Time 
				ans[idx] += alpha[targetStr[i] - 'A'][j]; //[s[i] - 'A'] The range of the number obtained is 0 To 26, Just corresponding to the letters ;[j] Represents the... Of the letter that will be read j There is a corresponding res in 
				if(targetStr[i + 1] <= 'Z' and targetStr[i + 1] >= 'A') ans[idx] += " "; // If the next letter in the target string is still a letter , Just add a space ( Because there are spaces between every letter in the result )
				++idx; //ans The subscript idx One back , Continue to access the same letter 7 The next line in the line 
			}
			f = true; // It means that the current is a letter , Mark the , Line breaks will work after reading a word 
		}
		else if(f == true){ // Can get here , Indicates that the current character is not a letter , It means that you have finished reading a word 
			f = false; // Reset 
			t += 7;  //t+=7 representative res Move the subscript back 7 position , Equivalent to reading a word and then wrapping 
		}
		idx = idx % 7 + t;  // The same word is on the same line , therefore idx The subscript of must be remainder , Otherwise, when outputting, the letters of the same word will run down 
	}
	
	for(int i = 0; ans[i].size() != 0; i++){ //ans Output line by line from the beginning , Until you encounter empty 
		if(i % 7 == 0 and i != 0) cout << "\n";  // Read a line of words , To wrap 
		cout << ans[i] << "\n";	// Imagine being a printer , Output line by line from top to bottom , So it's a new line 
	}
	
	return 0;
}