# Codeforces: 61 E. Enemy is weak

Solution Idea:

In this problem we need to count the number of triple inverse triplets on a given array. A triple inverse triplet is three positions i, j, k on an array A=[] such that i < j  Aj > Ak.

Now to solve this problem we need some range update and query data structure like segment tree or binary indexed tree. The idea is at first compress the array by mapping the numbers from range 1 to 10^6. After that read the array from reverse order and query the value of that mapped index on segment tree(let this mapped index is id)and store it on an array let name this array ara2=[]. After that increment the value of the index id+1 to n. It means that when we read any number from id+1 to n we can know that current number id have impact on this number. Here id is the k_th element of the triplet and all the number from id+1 to n is the j_th element of the triplet. Now each ara2[] index hold the number of inverse pair starting at id.

Now read the array element from the reverse order again and do the same thing. This time we need another segment tree. This time when we stand on position id add the value of the segment tree of position id to the answer and update all position in the range id+1 to n by ara2[index_of_id]. Do this for the whole array.

We can repeat this technique for 4,5,6 element inversion too.

```
#include &lt;bits/stdc++.h&gt;
#include &lt;ext/pb_ds/assoc_container.hpp&gt;
#include &lt;ext/pb_ds/tree_policy.hpp&gt;

#define pii              pair &lt;int,int&gt;
#define pll              pair &lt;long long,long long&gt;
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cerr&lt;&lt;#x &quot; = &quot;&lt;&lt;(x)&lt;&lt;endl
#define VI               vector &lt;int&gt;
#define DBG              pf(&quot;Hi\n&quot;)
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf(&quot;%d&quot;,&amp;a)
#define sfl(a)           scanf(&quot;%lld&quot;,&amp;a)
#define sff(a,b)         scanf(&quot;%d %d&quot;,&amp;a,&amp;b)
#define sffl(a,b)        scanf(&quot;%lld %lld&quot;,&amp;a,&amp;b)
#define sfff(a,b,c)      scanf(&quot;%d %d %d&quot;,&amp;a,&amp;b,&amp;c)
#define sfffl(a,b,c)     scanf(&quot;%lld %lld %lld&quot;,&amp;a,&amp;b,&amp;c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i&lt;n;i++)
#define loop1(i,n)       for(int i=1;i&lt;=n;i++)
#define REP(i,a,b)       for(int i=a;i&lt;b;i++)
#define RREP(i,a,b)      for(int i=a;i&gt;=b;i--)
#define TEST_CASE(t)     for(int z=1;z&lt;=t;z++)
#define PRINT_CASE       printf(&quot;Case %d: &quot;,z)
#define LINE_PRINT_CASE  printf(&quot;Case %d:\n&quot;,z)
#define CASE_PRINT       cout&lt;&lt;&quot;Case &quot;&lt;&lt;z&lt;&lt;&quot;: &quot;
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1&lt;&lt;28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;

//using namespace __gnu_pbds;
//typedef tree&lt;int, null_type, less&lt;int&gt;, rb_tree_tag, tree_order_statistics_node_update&gt; ordered_set;

/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

//int Set(int N,int pos){return N=N | (1&lt;&lt;pos);}
//int reset(int N,int pos){return N= N &amp; ~(1&lt;&lt;pos);}
//bool check(int N,int pos){return (bool)(N &amp; (1&lt;&lt;pos));}
/*------------------------------------------------*/

#define mx 1000006

ll tree[2][mx];
ll ara[mx],ara2[mx];

void update(int id, int idx, ll val)
{
for(; idx&lt;mx &amp;&amp; idx; idx+=idx&amp;-idx)
tree[id][idx]+=val;
}

ll query(int id, int idx)
{
ll ret=0;
for(; idx; idx-=idx&amp;-idx)
ret+=tree[id][idx];
return ret;
}

vector&lt;ll&gt;v;

int main()
{

//    freopen(&quot;in.txt&quot;,&quot;r&quot;,stdin);
//	  freopen(&quot;out.txt&quot;,&quot;w&quot;,stdout);

int n;
sf(n);
v.pb(0);
loop1(i,n)
{
sfl(ara[i]);
v.pb(ara[i]);
}
sort(all(v));

ll ans=0;

for(int i=n; i&gt;0; i--)
{
int id=lower_bound(all(v),ara[i])-v.begin();
ara2[i]=query(0,id);
update(0,id+1,1);
//        update(0,id-1,-1);
}

for(int i=n; i&gt;0; i--)
{
int id=lower_bound(all(v),ara[i])-v.begin();
ans+=query(1,id);
update(1,id+1,ara2[i]);
//        update(1,id-1,-ara2[i]);
}

printf(&quot;%lld\n&quot;,ans);

return 0;
}

```