九度 题目1550：分糖果

题目：http://ac.jobdu.com/problem.php?pid=1550

思路：两遍扫描。第一遍从i ==1~ i==N-2, 第二遍i ==N-2~ i==1 。每次拿出3个相邻元素a[i-1],a[i],a[i+1]进行考察，更新。先更新b[i],再更新b[i-1] ,b[i+1] ;

          此时：N==1或者N==2特判！

代码：

#include<cstdio>
#include<cstdlib>
#include<cstring>
int a[100001],b[100001];
int main()
{
	  int n,i;
          while(scanf("%d",&n)!=EOF)
	 {
	   for(i=0;i<n;i++)
	   {
		   scanf("%d",&a[i]);
		   b[i]=1;
	   }
	   if(n==1){ printf("1\n");continue;}
	   if(n==2) 
	   {
	      if(a[0]==a[1]) printf("2\n");
		  else printf("3\n");
		  continue;
	   }
	   if(b[0]>b[1]) b[0]++;
	   for(i=1;i<n-1;i++)
	   {
	       if(a[i]>a[i-1]&&b[i]<=b[i-1]) b[i]=b[i-1]+1;
		   if(a[i]>a[i+1]&&b[i]<=b[i+1]) b[i]=b[i+1]+1;
		   if(a[i-1]>a[i]&&b[i-1]<=b[i]) b[i-1]=b[i]+1;
		   if(a[i+1]>a[i]&&b[i+1]<=b[i]) b[i+1]=b[i]+1;
	   }
	   for(i=n-2;i>=1;i--)
	   {
	       if(a[i]>a[i-1]&&b[i]<=b[i-1]) b[i]=b[i-1]+1;
		   if(a[i]>a[i+1]&&b[i]<=b[i+1]) b[i]=b[i+1]+1;
		   if(a[i-1]>a[i]&&b[i-1]<=b[i]) b[i-1]=b[i]+1;
		   if(a[i+1]>a[i]&&b[i+1]<=b[i]) b[i+1]=b[i]+1;
	   }
	   long long sum=0;
	   for(i=0;i<n;i++)
		   sum+=b[i];
	   printf("%ld\n",sum);
	}
	return 0;
}