Equilibrium point

Given an array A your task is to tell at which position the equilibrium first occurs in the array. Equilibrium position in an array is a position such that the sum of elements below it is equal to the sum of elements after it.
Input:
The first line of input contains an integer T denoting the no of test cases then T test cases follow. First line of each test case contains an integer N denoting the size of the array. Then in the next line are N space separated values of the array A.
Output:
For each test case in a new  line print the position at which the elements are at equilibrium if no equilibrium point exists print -1.
Constraints:
1<=T<=100
1<=N<=100
Example:
Input:
2
1
1
5
1 3 5 2 2
Output:
1
3
Explanation:
1. Since its the only element hence its the only equilibrium point
2. For second test case equilibrium point is at position 3 as elements below it (1+3) = elements after it (2+2)

这个题目的题意就是找到一个平衡点，它之前的所有元素和等于它之后的所有元素和，如果不存在就输出“-1”。

解题思路：

首先算出数组总和，然后从第一个元素开始假设它是“当前平衡点”，计算其左右之和是否相等，如果相等，则输出该平衡点。否则假设下一个元素是“当前平衡点”进行依次遍历。

巧妙的是，我选择suml当作平衡点左边的元素之和，sumr起初是表示数组全部元素之和，但是在程序中可以利用它每次减去“当前平衡点数值大小”用来表示“当前平衡点右边元素之和”。然后对比即可，使得整个程序的时间复杂度得到了很好的控制。

下面是我的代码实现：

#include <iostream>
using namespace std;
int main()
{
    int T;cin>>T;
    int N;
    int *arr;
    while(T--)
    {
        cin>>N;
        arr=new int[N];
        int i,j,suml=0,sumr=0,flag=-1;
        for(i=0;i<N;i++)
        {
            cin>>arr[i];
            sumr+=arr[i];
        }
        for(i=0;i<N;i++)
        {
            sumr=sumr-arr[i];
            if(suml==sumr)
            {
                cout<<i+1<<endl;
                flag=1;
            }
            suml+=arr[i];
        }
        if(flag==-1)
            cout<<"-1"<<endl;
    }
    return 0;
}

　　如果有问题，欢迎留言哈。