二分查找与二分答案（待补充）

int mid=(left+right)/2;
//会有超过int类型的可能性
int mid=left+(right-left)/2;
//这么做可以避免运算溢出

int search(int x)
{
    int left = 1, right = n;
    while (left <= right)
    {
        int mid = left+( right-left) / 2;
        if (a[mid] == x)    return eeee(mid);
        else if (a[mid] > x)    right = mid - 1;
        else left = mid + 1;
    }
    return -1;

}