Solution: Compression Technology (original version and sequel version)

compression technique

Title Description

Let a Chinese character be composed of N × N Of 0 and 1 The dot matrix pattern of .

We generate compression codes according to the following rules . A continuous set of values ： Start from the first symbol in the first line of the Chinese character dot matrix pattern , In writing order from left to right , From top to bottom . The first number represents several consecutive 0, The second number means that there are several consecutive 1, The third number is followed by several consecutive 0, The fourth number is followed by several consecutive 1, And so on ……

for example : The following Chinese character dot matrix pattern ：

0001000
0001000
0001111
0001000
0001000
0001000
1111111

The corresponding compression code is ： 7 3 1 6 1 6 4 3 1 6 1 6 1 3 7 （ The first number is N , The rest of you say alternately 0 and 1 The number of , The compression code guarantees N × N= The sum of alternating numbers ）

Input format

a line , Compression code .

Output format

Chinese character dot matrix （ There is no space between dot matrix symbols ）.（3<=N<=200）

Examples #1

The sample input #1

7 3 1 6 1 6 4 3 1 6 1 6 1 3 7

Sample output #1

0001000
0001000
0001111
0001000
0001000
0001000
1111111

Answer key

At that time, I saw this problem , Think ： Ah, it's simple , Direct output is not the end .
Do it , People are stupid . So never think of a simple problem ！

sometimes , It's really “ Aware that , Lost thousands of miles ”. I haven't been right three times before , Just change one sentence of the program AC.

// Author:PanDaoxi
#include <iostream>
using namespace std;
//  You don't need a two-dimensional array , Array directly once 
int a[40201];
int main(){
    
 	int n,m,k=0,text,tmp=1,t;
 	// m*m Of the lattice 
	cin>>m;
	while(cin>>n){
    
		//  Characters used 
		text=k%2;
		//  Next , Connect n individual text
		t=tmp,tmp+=n;
		for(int i=t;i<tmp;i++){
    
			//  Write array 
			a[i]=text;
		}
		//  The next number 
		k++;
	}
	//  Output 
	for(int i=1;i<=m*m;i++){
    
		cout<<a[i];
		//  If the current number of characters meets the requirement of m to be divisible by , It's time for a new line 
		if(i%m==0) cout<<endl;
	}
	return 0;
}

You can use Boolean arrays , Further optimization ：

// Author:PanDaoxi
#include <iostream>
using namespace std;
bool a[40201],text=true;
int main(){
    
 	int n,m,tmp=1,t;
	cin>>m;
	while(cin>>n){
    
		text=!text,t=tmp,tmp+=n;
		for(int i=t;i<tmp;i++){
    
			a[i]=text;
		}
	}
	for(int i=1;i<=m*m;i++){
    
		cout<<a[i];
		if(i%m==0) cout<<endl;
	}
	return 0;
}

compression technique （ Sequel Edition ）

Title Description

Let a Chinese character be composed of N × N Of 0 and 1 The dot matrix pattern of .

We generate compression codes according to the following rules . A continuous set of values ： Start from the first symbol in the first line of the Chinese character dot matrix pattern , In writing order from left to right , From top to bottom . The first number represents several consecutive 0, The second number means that there are several consecutive 1, The third number is followed by several consecutive 0, The fourth number is followed by several consecutive 1, And so on ……

for example : The following Chinese character dot matrix pattern ：

0001000
0001000
0001111
0001000
0001000
0001000
1111111

The corresponding compression code is ： 7 3 1 6 1 6 4 3 1 6 1 6 1 3 7 （ The first number is N , The rest of you say alternately 0 and 1 The number of , The compression code guarantees N×N = The sum of alternating numbers ）

Input format

Chinese character dot matrix （ There is no space between dot matrix symbols ）.（3<=N<=200）

Output format

a line , Compression code .

Examples #1

The sample input #1

0001000
0001000
0001111
0001000
0001000
0001000
1111111

Sample output #1

7 3 1 6 1 6 4 3 1 6 1 6 1 3 7

Answer key

It took a lot of effort , Finally got it （ I can't ）
Think of a lot of ways , Thought of a strange way

// Author:PanDaoxi
#include <iostream>
#include <cmath>
using namespace std;
int a[40200],n=0,point=0; // point Save subscript 
int main(){
    
	char x,y='0';
	while(cin>>x){
     //  The input character 
		//  Number of characters +1
		n++;
		//  Judge ： If the present one is the same as the previous one 
		if(x==y) a[point]++;
		//  If it's not the same , Update character 
		else a[point++]++,y=x;
	}
	//  because n*n Matrix , that n= n The arithmetic square root of 
	cout<<sqrt(n)<<" ";
    //  Special value processing 
	a[0]--,a[point]++;
	//  The output is over 
	for(int i=0;i<=point;i++){
    
		cout<<a[i]<<" ";
	}
	return 0;
}