cf 61 E. Enemy is weak 离散化＋树状数组

题意：

给出一个数组，数组的每一个元素都是不一样的，求出对于３个数组下标 i, j, k such that i < j < k and a_i > a_j > a_k where a_x is the value at position x. 　的个数

明显数组的值太大了

先离散化，然后就是简单的树状数组了

对于每一个ｉ，只要统计ｉ前面的数中比a[i]大的数的个数，和ｉ后面的数中比a[i]小的数的个数即可

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <math.h>

#define LL long long

using namespace std;

const int maxn =  + ;

struct Node
{
    int id,init,chg;
};
Node node[maxn];
int fir[maxn];
int sec[maxn];
int c[maxn];

bool cmp1(Node x,Node y)
{
    return x.init<y.init;
}

bool cmp2(Node x,Node y)
{
    return x.id<y.id;
}

inline int lb(int x)
{
    return x & (-x);
}

void update(int x,int add)
{
    while(x <= maxn){
        c[x] += add;
        x += lb(x);
    }
}

int query(int x)
{
    int ret = ;
    while(x > ){
        ret += c[x];
        x -= lb(x);
    }
    return ret;
}

LL solve(int n)
{
    sort(node+,node+n+,cmp1);
    for(int i=;i<=n;i++)
        node[i].chg = i;
    sort(node+,node+n+,cmp2);
    memset(c,,sizeof c);
    for(int i=;i<=n;i++){
        fir[i] = i -  - query(node[i].chg - );
        update(node[i].chg,);
    }
    memset(c,,sizeof c);
    for(int i=n;i>;i--){
        sec[i] = query(node[i].chg - );
        update(node[i].chg,);
    }
    LL ret = ;
    for(int i=;i<=n;i++)
        ret += (LL) fir[i] * sec[i];
    return ret;
}

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=;i<=n;i++){
        scanf("%d",&node[i].init);
        node[i].id = i;
    }
    //printf("%I64d\n",solve(n));
    cout<<solve(n)<<endl;
    return ;
}