# Program to generate Look and Say series

#### This program will Generate the ‘Look and Say’ series for the given no.of lines from the User.

## Logic To Generate the Series :

- Get the No.of lines to print from the User to variable n .
- Initialise array a with 1 for first iteration .
- Using array b count the no.of occurences of 1 before any other number and insert the count value to even index of array b and insert the actual number in next position of array b.
- Continue this until array a reaches its end.
- Now print the array a values.
- Then copy array b to array a and repeat the steps for given no.of times from the user .

## Program :

#include<iostream> using namespace std; int main() { int n; //n=no.of lines to print cout<<"Enter the Number : "; cin>>n; int a[100*100*n]; //a[]=array to print series int b[100*100*n]; //b[]=array to store counts with actual numbers a[0]=1; //1 is fixed for first time int i,j,count,ia,ib; //count=stores count,ia=index of a[],ib=index of b[] ia=1; //ia=1 -> first time a[] contains only one element //iterate for n number of times to print the series for(int m=0;m<n;m++) { ib=0; //ib=0 -> for every iteration b[] is used to store values from beginning count=0; //count is made 0 after each line for(i=0,j=0;j<ia;j++) //loop which runs upto no.of element in a[] { if(a[i]==a[j]) //if condition executes if same number is found and increment count of that number { count++; } else //if different number is found else part gets executed { b[ib]=count; //at first b[] stores the count b[ib+1]=a[j-1]; //next b[] stores the actual number ib=ib+2; //since 2 positions of b[] is used so 2 times ib is incremented i=j; //now i stores the index of next different number to continue j=j-1; //j is decremented once to hold the current numbers' position count=0; //count is resetted in order to store the count of next number } } for(int k=0;k<ia;k++) //loop to print a[] { cout<<a[k]; } b[ib]=count; //the last numbers' count in a[] is stored here b[ib+1]=a[j-1]; //the last actual number in a[] is stored here ib=ib+2; //ib is incremented +2 cout<<"\n"; //one line of series is printed so go to new line is printed for(int k=0;k<ib;k++) //loop to store b[] in a[] to continue next iteration { a[k]=b[k]; } ia=ib; //ia now becomes no.of elements of ib } return 0; }