左右相等

Description

Once Bob took a paper stripe of n squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?

Input

The first input line contains integer n (1 ≤ n ≤ 10⁵) — amount of squares in the stripe. The second line contains n space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.

Output

Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.

Sample Input

Input

9 
1 5 -6 7 9 -16 0 -2 2

Output

Input

3 
1 1 1

Output

Input

2 
0 0

Output

1

分析：
    直接用循环分成两半进行比较

 1 #include<iostream>
 2 using namespace std;
 3 #define maxn 100000+5
 4 int n, ans,flag;
 5 int sum,lsum, rsum;
 6 int arr[maxn];
 7 int judge(int m)
 8 {
 9     lsum = 0, rsum = 0;
10     for (int i = 0; i <m; i++)
11     {
12         if (sum == 0)
13         {
14             if (i == m - 1)         //防止总和为零的时候再计算一次
15                 break;
16         }
17             lsum += arr[i];           //左边的数的和
18             rsum = sum - lsum;             //右边的数的和
19             if (lsum == rsum)             //相等就加加
20                 ans++;
21     }
22     if (ans)
23     {
24         flag = 1;
25         return flag;
26     }
27     else
28         return 0;
29 }
30 int main()
31 {
32      sum = 0,flag=0,ans=0;
33     cin >> n;
34     for (int i = 0; i < n; i++)
35     {
36         cin >> arr[i];
37         sum += arr[i];
38     }
39     if (sum % 2 != 0)
40         flag = 0;                  //总和为奇数，直接判断不可以分成相等的两半
41     else
42         judge(n);
43     if (flag == 1)
44         cout <<ans << endl;
45     else
46         cout << "0" << endl;
47 
48     return 0;
49 }