codeforece 18c

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

91 5 -6 7 9 -16 0 -2 2

Output

3

Input

31 1 1

Output

0

Input

20 0

Output

1

題意:

一張紙上畫着n個格子，每個格子上寫着一個整數（可能爲負整數），把這些格子分成兩段，使得這兩段之和相等，求出一共有多少種方案。

思路：

在輸入n個數的時候，用一個數組保存前i段之和，輸入完畢之後，求出其總和，再用一個循環，看前i段之和是不是總和的兩倍，如果是，那麼ans++，循環完畢之後，ans就是答案。要注意的是要用long long 保存，否則會數據溢出。

代碼：

#include<cstdio>
int  a[100010];
long long   b[100010];
using namespace std;
int main()
{
    long long ans;
    long long  sum;
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;
        sum=0;
        for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        sum+=a[i];
        b[i]=sum;
    //   printf("b[%d]%d\n",i,b[i]);

    }
  //  printf("%d\n",sum);
    for(int i=1;i<n;i++)

        if(2*b[i]==sum)
            ans++;

   printf("%I64d\n",ans);
    }
    return 0;
}