Approach

Let ~p_i~ be the prefix sum of the first ~i~ elements, with ~p_0 = 0~. The sum of the subarray from ~l~ to ~r~ is ~p_r - p_{l-1}~.

So when we are at a prefix sum ~p_r~, we need to know how many earlier prefix sums equal ~p_r - x~. Every such earlier prefix sum gives one subarray ending at ~r~ whose sum is ~x~.

We scan the array from left to right, maintain the current prefix sum, and store how many times each prefix sum has appeared in a dictionary.

This is ~O(n)~ time.

Solution (Python)

import sys


data = list(map(int, sys.stdin.buffer.read().split()))
n, x = data[0], data[1]
values = data[2 : 2 + n]

seen = {0: 1}
prefix = 0
answer = 0

for value in values:
    prefix += value
    answer += seen.get(prefix - x, 0)
    seen[prefix] = seen.get(prefix, 0) + 1

print(answer)