http://codeforces.com/problemset/problem/61/E

题意是求 i<j<k && a[i]>a[j]>a[k] 的对数

会树状数组求逆序数的话，这个推一下就能出结果：
做法：
1、离散化，由于a[i]能够达到1e9

2、插入a[i]的时候，记录x[i]=i-sum(a[i]); a[i]之前比a[i]大的有x[i]个

3、插入完毕后，求a[i] 之后比a[i]小的数的个数y[i]

ans=segma(x[i]*y[i])   注意x[i]*y[i]会超出int  由于这wa了一次

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const int INF = 100000000;
const int MAXN  = 1e6+100;
ll x[MAXN],y[MAXN];
int a[MAXN],tmp[MAXN];
ll c[MAXN];
int n;inline int lowb(int x) {return x&(-x);}void update(int x, int d)
{while(x<=n){c[x]+=d;x+=lowb(x);}
}ll sum(int x)
{ll ret=0;while(x>0){ret+=c[x];x-=lowb(x);}return ret;
}bool cmp(const int i, const int j)
{return a[i]<a[j];
}void dis()
{sort(tmp+1,tmp+1+n,cmp);int tt=0,pre=a[tmp[1]]-1;for(int i=1;i<=n;i++){if(pre!=a[tmp[i]]){pre=a[tmp[i]];a[tmp[i]]=++tt;}elsea[tmp[i]]=tt;}
}int main()
{//IN("B.txt");ll ans=0;while(~scanf("%d",&n)){for(int i=1;i<=n;i++){scanf("%d",&a[i]);tmp[i]=i;}dis();//*for(int i=1;i<=n;i++){printf("a[%d]=%d\n",i,a[i]);}*//CL(c,0);ans=0;for(int i=1;i<=n;i++){update(a[i],1);x[i]=i-sum(a[i]);//a[i]之前比a[i]大的有`x[i]个y[i]=sum(a[i]-1);//a[i]之前比a[i]小的数}for(int i=1;i<=n;i++){y[i]=sum(a[i]-1)-y[i];//a[i] 之后比a[i]小的数的个数ans+=x[i]*y[i];}//* for(int i=1;i<=n;i++)printf("i=%d x=%d y=%d\n",i,x[i],y[i]);*//printf("%I64d\n",ans);}return 0;
}