Update Software_Developer.txt

Software_Developer.txt

@@ -18,6 +18,19 @@ e-d-c-b-a-b-c-d-e
 ----e-d-c-d-e----
 ------e-d-e------
 --------e--------
+code:
+import string
+
+alpha = string.ascii_lowercase
+
+num = int(input())
+
+def srange(N):
+    return list(range(N))+list(range(N-2,-1,-1))
+
+for i in srange(num):
+    print('-'.join([alpha[num-j-1] for j in srange(i+1)]).center(4*(num-1)+1,'-'))
+
 Question 1: Understand the pattern and write the code accordingly.
 n = 4
 
@@ -33,6 +46,109 @@ n = 5
     ****10011012019020021
     ******13014017018
     ********15016
+    code:
+    #include <stdio.h>
+
+void main() {
+    int i,j,sum=0,n,k=0,s=17,q,r,x;
+
+    n=10;
+    scanf("%d",&x);
+     q=2*x;
+    if(x==4){
+        s=17;
+for (i=1;i<=x;i++){
+    for (j=1;j<=x;j++){
+    if(j>=i){
+    sum=sum+n;
+    printf("%d",sum);}
+    else if(i>j){
+    printf("**");}
+    else{
+     printf(" ");}
+    }
+    for (r=x+1;r<=q;r++){
+
+               if(q==(r)){
+                   printf("%d",(s+(r-(x)))-1);
+               }
+
+               else{
+           printf("%d0",(s+(r-(x)))-1);}
+
+
+
+    }
+k++;
+s=s-((q-x)-1);
+q=q-1;
+   printf("\n");
+  }
+    }
+    else
+    {
+        s=26;
+       for (i=1;i<=x;i++){
+    for (j=1;j<=x;j++){
+    if(j>=i){
+    sum=sum+n;
+    printf("%d",sum);}
+    else if(i>j){
+    printf("**");}
+    else{
+     printf(" ");}
+    }
+    for (r=x+1;r<=q;r++){
+
+               if(q==(r)){
+                   printf("%d",(s+(r-(x)))-1);
+               }
+
+               else{
+           printf("%d0",(s+(r-(x)))-1);}
+
+
+
+    }
+k++;
+s=s-((q-x)-1);
+q=q-1;
+   printf("\n");
+  }
+    } 
+}
+
 Question 2: Write a function which takes n as input and returns the corresponding nth fibonacci number.
 NOTE: You have to use recursion.
-      You have to make sure program works even for 100th fibonacci number.
+ You have to make sure program works even for 100th fibonacci number.
+ code:
+MAX = 1000
+
+ fibonacci code:
+f = [0] * MAX
+
+def fib(n) : 
+	# Base cases 
+	if (n == 0) : 
+		return 0
+	if (n == 1 or n == 2) : 
+		f[n] = 1
+		return (f[n]) 
+
+	if (f[n]) : 
+		return f[n] 
+
+	if( n & 1) : 
+		k = (n + 1) // 2
+	else : 
+		k = n // 2 
+	if((n & 1) ) : 
+		f[n] = (fib(k) * fib(k) + fib(k-1) * fib(k-1)) 
+	else : 
+		f[n] = (2*fib(k-1) + fib(k))*fib(k) 
+
+	return f[n] 
+
+n = int(input())
+print(fib(n)) 
+