The minimum unique array sum that can be achieved is when all elements in the array are different, resulting in the sum of the array being equal to the sum of the first n natural numbers, which is n(n1)/2.
To show that the solution of the recurrence relation t(n) t(n-1) n is in O(n2), we can use the Master Theorem. This theorem helps analyze the time complexity of recursive algorithms. In this case, the recurrence relation can be seen as T(n) T(n-1) n, which falls under the Master Theorem's first case where a 1, b 1, and f(n) n. Since f(n) n is polynomially larger than nlogb(a) n0, the solution is in O(n2).
The time complexity of the algorithm represented by the recurrence relation t(n) 4t(n/2) n2 logn is O(n2 log2 n).
The function t(n) relates to the function t(n1/2) 1 by taking the square root of n in the second function and adding 1 to the result.
The total running time of counting from 1 to n in binary is log(n1) steps.
P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n
Fibonacci numbers are the numbers in a sequence defined as follows: N1 = 1 N2 = 1 and after that, each number is the sum of the last two numbers in the sequence. N3 = N1 + N2 N4 = N2 + N3 and so on.
n3 + 1 = n3 + 13 = (n + 1)(n2 - n + 12) = (n + 1)(n2 - n + 1)
void main() { int i; float n1,n2; abc: printf("Enter two nos "); scanf("%f%f",&n1,&n2); printf("\n %f + %f = %f " ,n1,n2,n1+n2); printf("\n %f - %f = %f " ,n1,n2,n1-n2); printf("\n %f x %f = %f " ,n1,n2,n1*n2); printf("\n %f / %f = %f " ,n1,n2,n1/n2); printf("\npress 5 to make another calculation"); scanf("%d",&i); if (i==5) goto abc; }
It is a three-way case statement. ( Arithmetic IF statement) IF(exp) n1,n2,n3 meaning go to n1 if expr < 0 ( value is negative) go to n2 if expr = 0 ( value is zero) go to n3 if expr > 0 ( value is positive) IF(expr1-expr2) n1,n2,n3 meaning go to n1 if expr < expr2 go to n2 if expr = expr2 go to n3 if expr > expr2 reference: Fortran 77 with 90 Applications for scientific and engineers second edition RAMA N. REDDY CAROLA A. ZEIGLER
try a for loop: for x in 1 2 3 4 n=0 do n=`echo $x+$n`|bc done Integer arithmetic can be done in the shell itself without requiring an external program, making the operation much faster: n1=123 n2=543 sum=$(( $n1 + $n2 )) echo "$sum"
n x n2 = n3
They are reverse operations, in the sense that if y is the square of x then x is a square root of y. Remember though, that in the above scenario, -x is also a square root of y.
n3 + 3n2 + 4n + 12 = (n3 + 3n2) + (4n + 12) = n2(n + 3) + 4(n + 3) = (n2 + 4)(n + 3).
The formula for the sum of a series of cubes is as follows: 13 + 23 + 33 + ... + n3 = [n2*(n+1)2]/4 You may notice that this is the same as the square of the sum 1 + 2 + 3 + ... + n.
n3 - 93 can be factorised into (n - 9)(n2 + 9n + 81)
In the word "gallipolis"there are 10 letters (n =10)."l" appears 3 times (n1 =3)."i" appears 2 times (n2 =2).And 5 letters appear once (n3 =1, n4 =1, n5 =1, n6 =1, n7 =1)The number of permutations that can be made with these 10 letters is;P =n!/(n1!n2!n3!n4!n5!n6!n7!) =10!/(3!∙2!∙1!∙1!∙1!∙1!∙1!) =302 400