Fruit Ninja

本文介绍了一个关于寻找特定三元组数量的算法问题——Fruit Ninja。问题要求找到所有满足x<z<y条件的三元组(x, y, z)，并提供了一种有效的算法解决方案，利用了树状数组进行优化计算。

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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1675 Accepted Submission(s): 658

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

题意：问你 (x,y,z)有多少 x < z < y。

思路：直接算肯定是不行的，不过你可以用x<z<y的加上x<y<z的，再减去x<y<z的情况。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
long long a[100010];
long long mod=100000007;
long long lowbit(long long x)
{ return x&(-x);
}
long long sum(long long x)
{ long long ret=0;
  while(x)
  { ret+=a[x];
    x-=lowbit(x);
  }
  return ret;
}
void update(long long x)
{ while(x<=100005)
  { a[x]++;
    x+=lowbit(x);
  }
}
int main()
{ int T,t,n,m,i,j;
  long long k,ans,qx,qd,hx,hd;
  scanf("%d",&T);
  for(t=1;t<=T;t++)
  { memset(a,0,sizeof(a));
    ans=0;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    { scanf("%I64d",&k);
      qx=sum(k-1);
      qd=i-1-qx;
      hd=n-k-qd;
      ans-=qx*hd;
      ans+=hd*(hd-1)/2;
      update(k);
    }
    ans%=mod;
    printf("Case #%d: %I64d\n",t,ans);
  }
}