You are given an array of integers "array []" and an input "x". you have to find if there are any two numbers in your array such
that the sum of their square equals to x2 . in other words, you have to find out if there is any pair of i,j such that array[i]2 + array[j]2 = x 2 . if there are such pairs, you should print all such (i,j). otherwise print "there are no such pairs". example: array []: 6, -4, 6, 3, 9, 0, -1, -9, 2, 6 x: 5 you have 1 pair: (-4) 2 + (3) 2 = 5 2 so, your output would be corresponding array indices: (1, 3) note: the array size is 10 and you have to get the elements of the array from
Given an integer array a, size n, and an input integer value x.. Need to find all pairs of members a[i], a[j] such that (a[i],a[j],x) form a Pythagoras triplet, with x>a[i], a[j].
We need a double loop, the outer loop for a[i], and the inner loop finds a[j]. Here's a pseudocode.
int i, j, x, i2, x2, count=0; input x; x2=x*x; // economizing on cycles if x<0 {x=-x}; for i=1 to n-1 { if x>=i { // skip processing the impossible i2=i*i; // economizing on cycles for j=2 to n { if x>=j { // skip processing the impossible if i2+j*j==x2 { count++; print(i,j); } } } } } if count==0 { print("there are no matched pairs");
It will have a similar complexity as a bubble sort, n^2/2.
Tim spends 1/3 each weekday sleeping and 7/24 in school. We can write 1/3 as 8/24 so we have a common denominator. Now we can see that Tim sleeps for 1/24 time of a weekday more then he spends in school.
