UVALive 4329 Ping pong

树状数组。考虑ai（从0开始，则i左边共i个，右边n-i-1个），左边有x个比他大的，i-x个比他小的，右边有y个比他大的，n-i-1-y个比他大的。交叉乘一下就得到了以ai为裁判的比赛总数。把所有人都枚举一遍，加在一起就是答案，会超int。

如何才能知道ai左边有多少比他小的呢？假如aj<ai且j<i，用另一个数组b来标记这个数有没有出现过，那么b[aj] = 1;这样，比ai小的数的个数，就是b的前缀和。

 

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<vector>

#include<string>

#include<queue>

#include<cmath>

///LOOP

#define REP(i, n) for(int i = 0; i < n; i++)

#define FF(i, a, b) for(int i = a; i < b; i++)

#define FFF(i, a, b) for(int i = a; i <= b; i++)

#define FD(i, a, b) for(int i = a - 1; i >= b; i--)

#define FDD(i, a, b) for(int i = a; i >= b; i--)

///INPUT

#define RI(n) scanf("%d", &n)

#define RII(n, m) scanf("%d%d", &n, &m)

#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)

#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)

#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)

#define RFI(n) scanf("%lf", &n)

#define RFII(n, m) scanf("%lf%lf", &n, &m)

#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)

#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)

#define RS(s) scanf("%s", s)

///OUTPUT

#define PN printf("\n")

#define PI(n) printf("%d\n", n)

#define PIS(n) printf("%d ", n)

#define PS(s) printf("%s\n", s)

#define PSS(s) printf("%s ", n)

///OTHER

#define PB(x) push_back(x)

#define CLR(a, b) memset(a, b, sizeof(a))

#define CPY(a, b) memcpy(a, b, sizeof(b))

#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}



using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int MOD = 100000000;

const int INFI = 1e9 * 2;

const LL LINFI = 1e17;

const double eps = 1e-6;

const double pi = acos(-1.0);

const int N = 111111;

const int M = 22;

const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};



int a[N], b[N], c[N], n;



void update(int x)

{

    while(x <= N)

    {

        b[x]++;

        x += x & (-x);

    }

}



int sum(int x)

{

    LL s = 0;

    while(x)

    {

        s += b[x];

        x -= x & (-x);

    }

    return s;

}



int main()

{

    //freopen("input.txt", "r", stdin);

    //freopen("output.txt", "w", stdout);



    int t, k;

    LL ans;

    RI(t);

    while(t--)

    {

        RI(n);

        CLR(b, 0);

        REP(i, n)

        {

            RI(a[i]);

            c[i] = sum(a[i]);

            update(a[i]);

        }

        CLR(b, 0);

        ans = 0;

        FD(i, n, 0)

        {

            k = sum(a[i]);

            ans += LL(c[i]) * LL(n - i - 1 - k) + LL(i - c[i]) * LL(k);

            update(a[i]);

        }

        printf("%lld\n", ans);

    }

    return 0;

}