Hdu 4000 树状数组

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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;
}


SD--根据订单创建出库单(相关的函数列表的介绍系列篇(2)) 创建订单样例代码参见:http://blog.youkuaiyun.com/CompassButton/archive/2007/05/09/1601712.aspx 可以根据一个订单或者部分订单项来创建出库单,代码如下:* Populating the item data lx_sales_item-ref_doc = l_vbeln. lx_sales_item-ref_item = c_000010 阅读详情

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题意:给你一个长度为n的全排列,求有多少组(x,y,z)满足,x&lt;y&lt;z,且a【x】&lt;a【z】&lt;a【y】。 先求出满足,a【x】&lt;a【z】,a【x】&lt;a【y】有多少组, 对于当前点,后面有k个数比他大,那么有 k*(k-1)/2组。 我们在减去  a【x】&lt;a【y】&lt;a【z】的情况。 对于当前点,前面有k个数比他小,后面有u个数比他大,那么就...

胖亚亚的小酒馆 140

HDU4000 Fruit Ninja(树状数组)

嗯……这题是个标题党……【妈蛋其实大部分题目都是标题党啊我发现。。。

丛九九小朋友的NC博客 883

HDU 4000 Fruit Ninja 【树状数组】

题意:给1-N的一个排列,找出所有的(x, y, z) 使得x 题解:对于每个x,只需要找到比x靠后,而且比x大的数的个数,这样的组合数为:n*(n-1)/2 这里面有重复的情况:i 树状数组统计x之前出现的比x小的个数,剩下的就全部可以算出来了。 #include #include #include #include using namespace std

智障 1372

HDU4000Fruit Ninja【树状数组+组合数】

大意: 告诉你一个有n个数的序列 (1 -- n) 问其中有多少组 (a[i], a[j], a[k]) 满足i   分析: 这个题跟那个中间为峰值的题很像   我的第一反应为树状数组 求的就是在这个序列当中  小大中 一共会出现多少次 小大中 = (小中大 + 小大中) — 小中大 (小中大 + 小大中) 也就是说只要前面这个数是小的  后面的大的数中任取两个就可以了 用两个数

zhanzhaozhangxiaoyi的专栏 316

hdu4000 Fruit Ninja(树状数组)

题目戳这里题意:对1~n的一个排列,求所有的i<j<k且a[i]<a[k]<a[j]的个数。思路: 全局思路在于:小大中组合个数 = 小(大)(大)组合个数 - 小中大组合个数 小(大)(大)的意思是第一个元素比后两个元素要小的组合。 对于第i个数组元素,qian表示在i之前有几个比a[i]要小,hou表示i之后有几个比a[i]要大。 则小中大 组合个数= qian * hou; 小(大)

a novice 585

HDU 4000 Fruit Ninja(树状数组)

Fruit Ninja 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 ar

持之以恒 481

HDU-4000 Fruit Ninja 树状数组+思维

Fruit Ninja Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3634    Accepted Submission(s): 1109   Problem Description Recently, dobby is ad...

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