Tanya and Candies

题意翻译

$\texttt{Tanya}$ 有 $n$ 个糖果，每个糖果都有一个重量 $a$。 她打算吃掉其中的 $(n-1)$ 颗糖果并将剩下的一颗给她的 $\texttt{dad}$。她吃糖果必须按照编号递增的顺序来吃，每天吃一颗。 她的爸爸只会吃名叫 `good candy` 的糖果。如果第 $i$ 个糖果是 `good candy`，那么，$\texttt{Tanya}$ 在偶数天吃的糖果的重量等于 $\texttt{Tanya}$ 在奇数天吃的糖果的重量。 **注意：**$\texttt{Tanya}$ 首先会给爸爸糖，之后自己再吃 你的任务是去找 `good candy` 的数量。

题目描述

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 $ .

输入输出格式

输入格式

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.

输出格式

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.

输入输出样例

输入样例 #1

7
5 5 4 5 5 5 6

输出样例 #1

2

输入样例 #2

8
4 8 8 7 8 4 4 5

输出样例 #2

2

输入样例 #3

9
2 3 4 2 2 3 2 2 4

输出样例 #3

3

说明

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] $ .