Tanya and Candies
Contest: Codeforces Round 540 (Div. 3)
Difficulty rating: 1200
Tags: implementation

Tanya has $$$n$$$ candies numbered from $$$1$$$ to $$$n$$$. The $$$i$$$-th candy has the weight $$$a_i$$$.

She plans to eat exactly $$$n-1$$$ candies and give the remaining candy to her dad. Tanya eats candies in order of increasing their numbers, exactly one candy per day.

Your task is to find the number of such candies $$$i$$$ (let's call these candies good) that if dad gets the $$$i$$$-th candy then the sum of weights of candies Tanya eats in even days will be equal to the sum of weights of candies Tanya eats in odd days. Note that at first, she will give the candy, after it she will eat the remaining candies one by one.

For example, $$$n=4$$$ and weights are $$$[1, 4, 3, 3]$$$. Consider all possible cases to give a candy to dad:

- Tanya gives the $$$1$$$-st candy to dad ($$$a_1=1$$$), the remaining candies are $$$[4, 3, 3]$$$. She will eat $$$a_2=4$$$ in the first day, $$$a_3=3$$$ in the second day, $$$a_4=3$$$ in the third day. So in odd days she will eat $$$4+3=7$$$ and in even days she will eat $$$3$$$. Since $$$7 \ne 3$$$ this case shouldn't be counted to the answer (this candy isn't good).
- Tanya gives the $$$2$$$-nd candy to dad ($$$a_2=4$$$), the remaining candies are $$$[1, 3, 3]$$$. She will eat $$$a_1=1$$$ in the first day, $$$a_3=3$$$ in the second day, $$$a_4=3$$$ in the third day. So in odd days she will eat $$$1+3=4$$$ and in even days she will eat $$$3$$$. Since $$$4 \ne 3$$$ this case shouldn't be counted to the answer (this candy isn't good).
- Tanya gives the $$$3$$$-rd candy to dad ($$$a_3=3$$$), the remaining candies are $$$[1, 4, 3]$$$. She will eat $$$a_1=1$$$ in the first day, $$$a_2=4$$$ in the second day, $$$a_4=3$$$ in the third day. So in odd days she will eat $$$1+3=4$$$ and in even days she will eat $$$4$$$. Since $$$4 = 4$$$ this case should be counted to the answer (this candy is good).
- Tanya gives the $$$4$$$-th candy to dad ($$$a_4=3$$$), the remaining candies are $$$[1, 4, 3]$$$. She will eat $$$a_1=1$$$ in the first day, $$$a_2=4$$$ in the second day, $$$a_3=3$$$ in the third day. So in odd days she will eat $$$1+3=4$$$ and in even days she will eat $$$4$$$. Since $$$4 = 4$$$ this case should be counted to the answer (this candy is good).

In total there $$$2$$$ cases which should counted (these candies are good), so the answer is $$$2$$$.

Input format:
The first line of the input contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of candies.

The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^4$$$), where $$$a_i$$$ is the weight of the $$$i$$$-th candy.

Output format:
Print one integer — the number of such candies $$$i$$$ (good candies) that if dad gets the $$$i$$$-th candy then the sum of weights of candies Tanya eats in even days will be equal to the sum of weights of candies Tanya eats in odd days.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first example indices of good candies are $$$[1, 2]$$$.

In the second example indices of good candies are $$$[2, 3]$$$.

In the third example indices of good candies are $$$[4, 5, 9]$$$.

Example input:
7
5 5 4 5 5 5 6

Example output:
2

Example input:
8
4 8 8 7 8 4 4 5

Example output:
2

Editorial:
Let's maintain four variables $$$oddPref$$$, $$$evenPref$$$, $$$oddSuf$$$ and $$$evenSuf$$$, which will mean the sum of $$$a_i$$$ with odd $$$i$$$ on prefix, even $$$i$$$ on prefix, odd $$$i$$$ on suffix and even $$$i$$$ on suffix. Initially, $$$oddPref$$$ and $$$evenPref$$$ are $$$0$$$, $$$oddSuf$$$ equals to the sum of $$$a_i$$$ with odd $$$i$$$ in a whole array and $$$evenSuf$$$ equals to the sum of $$$a_i$$$ with even $$$i$$$ in a whole array.
Let's iterate from left to right over all elements of the array. Let's consider the current element $$$a_i$$$. If $$$i$$$ is even, then set $$$evenSuf := evenSuf - a_i$$$, otherwise let's set $$$oddSuf := oddSuf - a_i$$$. Then let's consider we give the current candy to dad. Then we have to increase the answer if $$$oddPref = evenSuf$$$ and $$$evenPref = oddSuf$$$. Then if $$$i$$$ is even then let's set $$$evenPref := evenPref + a_i$$$, otherwise let's set $$$oddPref := oddPref + a_i$$$.