Codeforces Beta Round #57 (Div. 2) E. Enemy is weak

求满足条件的三元组的个数，可以转换求一元组和二元组组成的满足条件的三元组的个数，且对于(x),(y,z),x > y,且x出现的p_x < p_y。

x可直接枚举O(n),此时需要往后查询二元组的个数，二元组可由两个一元组生成，实际上到这里就转换成了求两次逆序对而已。

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>
#include <set>
#include <map>
#include <list>
#include <queue>
#include <string>
#include <iostream>
#include <algorithm>
#include <functional>
#include <stack>
using namespace std;
typedef long long ll;
#define T int t_;Read(t_);while(t_--)
#define dight(chr) (chr>='0'&&chr<='9')
#define alpha(chr) (chr>='a'&&chr<='z')
#define INF (0x3f3f3f3f)
#define maxn (1000005)
#define hashmod 100000007
#define ull unsigned long long
#define repne(x,y,i) for(i=(x);i<(y);++i)
#define repe(x,y,i) for(i=(x);i<=(y);++i)
#define repde(x,y,i) for(i=(x);i>=(y);--i)
#define repdne(x,y,i) for(i=(x);i>(y);--i)
#define ri register int
void Read(int &n){char chr=getchar(),sign=;for(;!dight(chr);chr=getchar())if(chr=='-')sign=-;
    for(n=;dight(chr);chr=getchar())n=n*+chr-'';n*=sign;}
void Read(ll &n){char chr=getchar(),sign=;for(;!dight(chr);chr=getchar())if
    (chr=='-')sign=-;
    for(n=;dight(chr);chr=getchar())n=n*+chr-'';n*=sign;}
int n,a[maxn],b[maxn];
ll c[maxn],s[maxn];
void hasharray(){
    memcpy(b,a,n*sizeof(a[]));
    sort(b,b+n);
    for(int i = ;i < n;++i) a[i] = (int)(lower_bound(b,b+n,a[i]) - b) + ;
}
int lowbit(int x){return x & (-x);}
void update(int x,ll v){
    while(x <= n){
        s[x] += v;
        x += lowbit(x);
    }
}
ll query(int x){
    ll ans = ;
    while(x){
        ans += s[x];
        x -= lowbit(x);
    }
    return ans;
}
int main()
{
    freopen("a.in","r",stdin);
    freopen("b.out","w",stdout);
    Read(n);
    ri i;
    repne(,n,i) Read(a[i]);
    hasharray();
    update(a[n-],);
    repde(n-,,i) c[i] = query(a[i]-),update(a[i],);
    memset(s,,sizeof(s));
    update(a[n-],c[n-]);
    ll ans = ;
    repde(n-,,i) ans += query(a[i]-),update(a[i],c[i]);
    cout << ans << endl;
    return ;
}