HDU 4000 Fruit Ninja 树状数组 + 计数

给你N的一个排列，求满足：a[i] < a[k] < a[j] 并且i < j < k的三元组有多少个。

一步转化：

求出所有满足 a[i] < a[k] < a[j] 并且i < j < k的三元组 与 a[i] < a[k] < a[j] 并且i < k < j的三元组 的个数的总和，设high[i]为在 i 右侧且大于a[i] 的数的个数，上述总和即为：high[i] * (high[i] - 1) / 2。

设low[i]为在 i 左侧且小于a[i] 的数的个数, a[i] < a[k] < a[j] 并且i < k < j的三元组 的个数即为：low[i]*high[i]。

答案即为：∑( high[i] * (high[i] - 1) / 2 - low[i]*high[i] );

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

#define LL long long int

using namespace std;

const int MAXN = 100100;
const LL MOD = 100000007;

int N;
int C[MAXN];
int num[MAXN];
int low[MAXN];
int high[MAXN];

int lowbit( int x )
{
    return x & ( -x );
}

void Update( int x, int val )
{
    while ( x <= N )
    {
        C[x] += val;
        x += lowbit(x);
    }
    return;
}

int Query( int x )
{
    int res = 0;
    while ( x > 0 )
    {
        res += C[x];
        x -= lowbit(x);
    }
    return res;
}

int main()
{
    int T, cas = 0;
    scanf( "%d", &T );
    while ( T-- )
    {
        scanf( "%d", &N );
        for ( int i = 1; i <= N; ++i )
            scanf( "%d", &num[i] );

        memset( C, 0, sizeof(C) );

        for ( int i = 1; i <= N; ++i )
        {
            low[i] = Query( num[i] - 1 );
            Update( num[i], 1 );
        }

        memset( C, 0, sizeof(C) );
        for ( int i = N; i > 0; --i )
        {
            //printf("i=%d %d %d
", i, Query( N ), Query( num[i] ) );
            high[i] = Query( N ) - Query( num[i] );
            Update( num[i], 1 );
        }

        LL ans = 0;
        for ( int i = 1; i <= N; ++i )
        {
            //printf( "h=%d l=%d
", high[i], low[i] );
            ans += (LL)high[i] * ( high[i] - 1 ) / 2 - (LL)low[i] * high[i];
            ans = ( ans + MOD ) % MOD;
        }

        printf( "Case #%d: %I64d
", ++cas, ans % MOD );
    }
    return 0;
}