boj problem 1326 冒泡的話複雜度太高 TLE錯誤。輸入時就應該進行判斷

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為結果 //temp為存儲目前位置1的個數，隻要後面出現一個-1 那麼就要碰撞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;

}