HDU 4000 Fruit Ninja(ツリー配列+組み合わせの問題)

3177 ワード

ix>y(z)の個数を先に求め,x>y>zの個数を求める.(small[i]は現在の1-i-1元素間のa[i]より小さい個数を記録し、high[i]は現在のi+1-n元素がa[i]より大きい個数を記録する)、x>y>zの個数はsmall[i]*high[i]...である.
smallは木の配列で求めることができます!!!
标题:posx構想:樹状配列のテーマ.まず,問題条件を満たすすべてのグループ数xProblem 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

 
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n;
const int maxn=111111;
int a[maxn],b[maxn];
__int64 small[maxn],high[maxn];

void update(int x,int num)
{
	while(x<=n){
		a[x]+=num;
		x+=x&(-x);
	}
}

__int64 getsum(int x)
{
	__int64 s=0;
	while(x>0){
		s+=a[x];
		x-=x&(-x);
	}
	return s;
}

int main()
{
	int i,j,t;
	__int64 ans;
	while(scanf("%d",&t)!=EOF){
		for(int _=1;_<=t;_++){
			memset(a,0,sizeof(a));
			memset(b,0,sizeof(b));
			scanf("%d",&n);
	        ans=0;
	    	for(i=1;i<=n;i++){
     			scanf("%d",&b[i]);
    	 		small[i]=getsum(b[i]);
    			high[i]=n-b[i]-(i-1-small[i]);
    		    update(b[i],1);
    		}
    		for(i=1;i<=n;i++){
     			ans+=high[i]*(high[i]-1)/2-small[i]*high[i];
    		}
    		printf("Case #%d: %d
",_,ans%100000007); } } return 0; }