Extracting the numbers of two elements which their sum is equal to their multiplication-CodePudding

we have array A that has N len(N = the len of the array) I have to found the numbers of pairs (J, I) that work on the following statement:

A[J] A[I} == A[J] * A[I]

1<=i, j<=N

(1 ≤ N ≤ 40000)

(0 ≤ A(i) ≤ 10^9)

example:

input:

3
2 4 2

output:

well i counldn't know how to limit the input size to only have 2 spaces or to split it if there was any more than the N

edit*

A = []
N = int(input(""))
B = (input(""))

B = B.split()
z = 0
myList = []
mylist2= []
pairs = 0
for q in B:
    if z < N:
        myList.append(q)
        z  = 1
    elif z >= N:
        break

for w in myList:
    w = int(w)
    mylist2.append(w)

for i in mylist2:
    for k in mylist2:`enter code here`
        if i   k == i * k:
            pairs 1

that what i have done so far

CodePudding user response：

So, as already mentioned in comments only pairs (2, 2) and (0, 0) satisfy the condition. The number of (0, 0) pairs is count(0) * (count(0) - 1) / 2. The same for (2, 2) pairs. Expressing this in python (assuming that array a is given).

def countsumprod(a):
    c0 = a.count(0)
    c2 = a.count(2)
    return (c0 * (c0 - 1)   c2 * (c2 - 1)) // 2