boj problem 1326 冒泡的话复杂度太高 TLE错误。输入时就应该进行判断

Description
There are N boxes on one straight line which is long enough.
They move at the same speed, but their directions may be different. That is, some boxes
may move left, while the others move right. As a beautiful girl fond of algorithm and
programming, Alice finds that two boxes moving toward each other will collide, and
after collision their directions change while their speeds remain the same. Alice also
knows that the boxes will not collide any more after many times of collision, she names
this final status as the stable status. The task is to help her count the number of
collisions before reach the stable status.

Input
here are multiple test cases. For each test case, the first line is an integer N (1 <= N <= 10000) representing the number of boxes, and the second line is N integers,
separated by spaces. The i-th integer will be -1 if the i-th box move left, otherwise,
it will be 1.
An negative integer indicates the end of input and should not be processed by your program.

Output
For each test case, output an integer which is the number of collisions to reach the
stable status on a single line in the format as indicated.

Sample Input

3
1 -1 1
4
1 -1 1 -1
-1

Sample Output

Case 1: 1
Case 2: 3

#include<iostream>
using namespace std;
int main()
{

  int i,j,p,k=0;

  int a[10000],b[10000];
  for(i=0;i<10000;i++)
   b[i]=0;
 
  while(cin>>p)
  {
 
     if(p<0)
   break;
  if(p>0)
  {
   int res=0,temp=0;、、//res为结果 //temp为存储目前位置1的个数，只要后面出现一个-1 那么就要碰撞temp次
    k++;
    for(i=0;i<p;i++)
    {
      cin>>a[i];
   if(a[i]==1)
    temp++;            
   else
    res=res+temp;
    }
    b[k]=res;
 
        
   
  }
  }
  for(i=1;i<k+1;i++)
   cout<<"Case "<<i<<":"<<b[i]<<" "<<endl;
  return 0;
}