CF E. Enemy is weak 线段树

E. Enemy is weak

time limit per test
5 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

The Romans have attacked again. This time they are much more than the Persians but Shapur is ready to defeat them. He says: "A lion is never afraid of a hundred sheep".

Nevertheless Shapur has to find weaknesses in the Roman army to defeat them. So he gives the army a weakness number.

In Shapur's opinion the weakness of an army is equal to the number of triplets
i, j, k such that
i < j < k and ai > aj > ak where
ax is the power of man standing at position
x. The Roman army has one special trait — powers of all the people in it are distinct.

Help Shapur find out how weak the Romans are.

Input
The first line of input contains a single number n (3 ≤ n ≤ 106) — the number of men in Roman army. Next line contains
n different positive integers
ai (1 ≤ i ≤ n, 1 ≤ ai ≤ 109) — powers of men in the Roman army.

Output
A single integer number, the weakness of the Roman army.

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use
cout (also you may use
%I64d).

Sample test(s)

Input

3
3 2 1

Output

1

Input

3
2 3 1

Output

0

Input

410 8 3 1

Output

4

Input

41 5 4 3

Output

1思路：
big[i] 记录是比a[i] 大且位于a[i] 前面的数的个数；
little[i] 是记录比a[i] 小 且位于 a[i] 后面数的个数；
那么 big[i]个数中选一个，a[i], little[i]中选一个， 这样的组合肯定满足题目要求；
所以 ans= ans+ big[i]*1*little[i] (for i=0 到 n )
怎么最快的求big[i],little[i]呢？ 用线段树了；

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<list>
#include<map>
#include<set>
using namespace std;
const int N=1000010;
int n;
int a
;
int big
,little
;
int cnt[N*4];

map<int,int>m;
int all;
void init(){
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
big[i]=a[i];
}
sort(big,big+n);
all=unique(big,big+n)-big;
for(int i=1;i<=all;i++) m[big[i-1]]=i;
for(int i=0;i<n;i++) a[i]=m[a[i]];

}
int require(int x,int y,int st,int ed,int i){
if(x>ed || y<st) return 0;
if(x<=st && y>=ed) return cnt[i];
int mid=(st+ed)>>1;
return require(x,y,st,mid,i*2)+require(x,y,mid+1,ed,i*2+1);

}

void updata(int x,int st,int ed,int i){
if(x<st || x>ed) return ;
cnt[i]++;
if(st==ed) return ;
int mid=(st+ed)>>1;
updata(x,st,mid,i*2);
updata(x,mid+1,ed,i*2+1);

}
void deal_big(){
memset(cnt,0,sizeof(big));

for(int i=0;i<n;i++){
big[i]=require(a[i]+1,all,1,all,1);
updata(a[i],1,all,1);
}
//    for(int i=0;i<n;i++)
//        printf("big==%d\n",big[i]);
}
void deal_little(){
memset(cnt,0,sizeof(cnt));
for(int i=n-1;i>=0;i--){
little[i]=require(1,a[i]-1,1,all,1);
updata(a[i],1,all,1);
}
//    for(int i=0;i<n;i++)
//        printf("ll=%d\n",little[i]);
}
void work(){
long long ans=0;
for(int i=0;i<n;i++)
ans=ans+(1LL)*big[i]*little[i];
cout << ans << endl;
}
int main()
{

//     freopen("in.in","r",stdin);
init();
deal_big();
deal_little();
work();
return 0;
}