Hdu 4000 树状数组

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最新推荐文章于 2022-02-25 19:38:17 发布

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分类专栏： ACM 文章标签： permutation integer input each output c

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本文详细介绍了如何使用树状数组解决FruitNinja游戏中特定条件下的三元组计数问题，包括输入解析、树状数组操作及最终输出结果的方法。

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Fruit Ninja

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 19 Accepted Submission(s) : 11

Problem Description

Recently, dobby is addicted in the Fruit Ninja. As you know, dobby is a free elf, so unlike other elves, he could do whatever he wants.But the hands of the elves are somehow strange, so when he cuts the fruit, he can only make specific move of his hands. Moreover, he can only start his hand in point A, and then move to point B,then move to point C,and he must make sure that point A is the lowest, point B is the highest, and point C is in the middle. Another elf, Kreacher, is not interested in cutting fruits, but he is very interested in numbers.Now, he wonders, give you a permutation of 1 to N, how many triples that makes such a relationship can you find ? That is , how many (x,y,z) can you find such that x < z < y ?

Input

The first line contains a positive integer T(T <= 10), indicates the number of test cases.For each test case, the first line of input is a positive integer N(N <= 100,000), and the second line is a permutation of 1 to N.

Output

For each test case, ouput the number of triples as the sample below, you just need to output the result mod 100000007.

Sample Input

2
6
1 3 2 6 5 4
5 
3 5 2 4 1

Sample Output

Case #1: 10
Case #2: 1
 
 

判断满足i<j<k且num[i]<num[k]<num[j]的总组数利用树状数组可以求出一个数前面比它小的数的个数，进而可以知道前面比它大的数的个数，总的比它大的个数减去前面比它大的个数等于后面比它大的个数，所以答案就很明显了~
代码：
#include<stdio.h>
#include<string.h>
int c[100010];
int n;
int lowbit(int x)
{    
    return x&(-x);
}
void update(int x,int d)
{    
    while(x<=n)    
	{        
	  c[x]+=d;        
	  x+=lowbit(x);    
    }
}
int sum(int x)
{    
     int ans=0;    
	 while(x>0)    
	 {        
	   ans+=c[x];        
	   x-=lowbit(x);    
     }    
	 return ans;
}
int main()
{    
    __int64 ans;    
	int t,cases=1;    
	int i,a;    
	scanf("%d",&t);    
	while(t--)    
	{        
	    ans=0;        
		memset(c,0,sizeof(c));        
		scanf("%d",&n);        
		for(i=1;i<=n;i++)        
		  {            
		     scanf("%d",&a);           
			  update(a,1);            
			  int presmaller=sum(a-1);            
			  int totbigger=n-a;            
			  int prebigger=i-sum(a-1)-1;           
			   __int64 afterbigger=totbigger-prebigger;           
			    ans-=presmaller*afterbigger;            
				if(afterbigger>=2)                
				ans+=afterbigger*(afterbigger-1)/2;        
		  }        
		  printf("Case #%d: %I64d\n",cases++,ans%100000007);    
    }    
return 0;
}