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现排序，然后从n*m往下找，直到所有的数都被找过

Posted by kric at 2007-04-13 21:20:21 on Problem 1323

		for(i=n*m,j=0,t=0;j<n;i--)//p[j]降序排列，t记录比当前p[j]大的数的个数，cnt是结果
		{
          if(i==p[j])
		  {
			  if(t)
			  {
				  t--;
				  j++;
			  }
			  else
			  {
			       cnt++;
				   j++;
			  }
		  }
		  else
			  t++;
		}

Followed by:

Re:现排序，然后从n*m往下找，直到所有的数都被找过 (278) wuzhipingdixiao 2009-04-01 18:13:49
Re:现排序，然后从n*m往下找，直到所有的数都被找过 (34) wuzhipingdixiao 2009-04-01 18:14:47

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