Complexity of Time from O (n^3) to O (n^2) optimization

Title Description

A given array of n integers a[0],a[1],a[2],a[3],.... a[n-1]. You want to output a two-dimensional array of n*n B,
where array b[i,j] (I first method of thinking

For i =0,1,2,... n-2
    for j = i+1,i+2,.... n-1
        add a[i]->a[j] and assign to B[i,j]

Because the j>i is satisfied, the diagonal of the two-dimensional matrix does not need to be computed, and the last line is always j>=i and does not have to be computed, so the first layer for loop is [0~n-2]
The outer loop operand N level, the inner layer is summed with Gaussian n^2 level, so time complexity is O (n^3) The second solution

For i = 0,1,2...n-2
    b[i][i+1] = A[i] + a[i+1] for

i = 0,1,2,3...n-2 for
    j = i+2,i+3,.... n-1
        b[i][j] = b[i] [j] + A[j]

Ideas:
B[0][1] = a[0] + a[1]
B[0][2] = a[0] + a[1] + a[2] = b[0][1] + a[2]
B[0][3] = a[0] + a[1] + a[2] + a[3] = b[0][2] + a[3]
。。。

B-matrix B[i][j] = B[i][j-1] + a[j], so that the intermediate calculation results can be used, but also reduce the repetition of the calculation.
Parsing algorithm, first for loop O (n), second two for Loop O (n^2), so the time complexity of the entire algorithm is O (n^2) program code

"" "
given by n integers a[1],a[2],a[3],.... A[n] Array A. You want to output a two-dimensional array of n*n B,
where the array B[i,j] (I<J) contains an array of items a[i]~a[j] and, that is, and a[i]+a[i+1] ..... A[J] (as long as i>=j, the
value of the item b[i,j] of the array is not specified, so regardless of the output of these values.)
"" " def fun (a):
    n = Len (a)
    B = [[0 for I in range (0,n)] for J in Range (0,n)] for
    I in range (0,n-1): for
        J in R Ange (i+1,n):
            sum = (A[i]+a[j]) * (j-i+1)//2
            b[i][j] = sum
    return B

def fun2 (a):
    n = Len (a)
    B = [[0 for I in range (0,n)] for J in Range (0,n)] for
    I in range (0,n-1):
        b[i][i+1] = A[i] + a[i+1] for

    i in Range (0,n-1):
        for J in Range (I+2,n):
            b[i][j] = b[i][j-1] + a[j]
    return B


def test ():
    A = [A. 4,5]
    result = Fun (a) for
    I in range (len (A)):
        print (Result[i])
def test2 ():
    A = [1,2,3,4,5]
    result = Fun2 (a) for
    I in range (Len (a)):
        print (result[i])

if __name__  = = ' __main__ ':
    Test ()
    print ("=====")
    test2 ()

The program needs to be verified, if there are errors thank you for pointing