Suppose, there is an array 'nums' of size n containing positive integers. We have another array 'queries' that contain integer pairs (pi, qi). For every query in the array queries, the answer will be the sum of numbers in the array nums[j] where pi <= j < n and (j - pi) is divisible by qi. We have to return the answer of all such queries, and if it is a large value, we return the answer modulo 10^9 + 7.
So, if the input is like nums = [2, 3, 4, 5, 6, 7, 8, 9, 10], queries = [(2, 5), (7, 3), (6, 4)], then the output will be [13, 9, 8].
To solve this, we will follow these steps −
A := nums
Q := queries
n := length of nums
M := 10^9 + 7
m := integer value of (n ^ 0.5) + 2
P := a new list containing the list A m times
for i in range 1 to m, do
for j in range n-1 to -1, decrease by 1, do
if i+j < n, then
P[i, j] := (P[i, j]+P[i, i+j]) modulo M
for each value b, k in Q, do
if k < m, then
return [value from index P[k, b]]
otherwise
return [sum(A[b to k]) modulo M]
Example
Let us see the following implementation to get better understanding −
def solve(A, Q):
n, M = len(A), 10**9+7
m = int(n**0.5)+2
P = [A[:] for _ in range(m)]
for i in range(1,m):
for j in range(n-1,-1,-1):
if i+j < n:
P[i][j] = (P[i][j]+P[i][i+j]) % M
return [P[k][b] if k < m else sum(A[b::k]) % M for b, k in Q]
print(solve([2, 3, 4, 5, 6, 7, 8, 9, 10], [(2, 5), (7, 3), (6, 4)]))Input
[2, 3, 4, 5, 6, 7, 8, 9, 10], [(2, 5), (7, 3), (6, 4)]
Output
[13, 9, 8]