Codeforces #261 D

Codeforces #261 D

D. Pashmak and Parmida's problem

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Parmida is a clever girl and she wants to participate in Olympiads this year. Of course she wants her partner to be clever too (although he's not)! Parmida has prepared the following test problem for Pashmak.

There is a sequence a that consists of n integers a1, a2, ..., an. Let's denote f(l, r, x) the number of indices k such that: l ≤ k ≤ r andak = x. His task is to calculate the number of pairs of indicies i, j (1 ≤ i < j ≤ n) such that f(1, i, ai) > f(j, n, aj).

Help Pashmak with the test.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 106). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a single integer — the answer to the problem.

简单的树状数组求逆序数。。。开始怎么也没想到。。。

答案没用long long wa一次

 #include <cstring>

 #include <iostream>

 #include <algorithm>

 #include <cstdio>

 #include <cmath>

 #include <map>

 #include <cstdlib>

 #define M(a,b) memset(a,b,sizeof(a))

 using namespace std;

 int n;

 int num[];

 int savefro[],saveb[];

 int ans[];

 int flag[];

 map<int,int> front,back;

 int bit[],cnn;

 int sum(int i)

 {

     int res = ;

     while(i>)

     {

         res+=bit[i];

         i-=i&-i;

     }

     return res;

 }

 void add(int i,int x)

 {

     while(i<=cnn)

     {

         bit[i]+=x;

         i+=i&-i;

     }

 }

 int main()

 {

     while(scanf("%d",&n)==)

     {

         front.clear();

         back.clear();

         M(saveb,);

         M(savefro,);

         M(ans,);

         M(flag,);

         M(bit,);

         long long res = ;

         for(int i = ;i<n;i++)

         {

             scanf("%d",&num[i]);

         }

         for(int i = ;i<n;i++)

         {

             savefro[i] = ++front[num[i]];

         }

         cnn = n;

         for(int i = n-;i>=;i--)

         {

             saveb[i] = ++back[num[i]];

             add(saveb[i],);

             res+=sum(savefro[i-]-);

            // cout<<' '<<sum(savefro[i-1]-1)<<' '<<savefro[i-1]<<endl;

         }

         /*cout<<"savefro"<<endl;

         for(int i = 0;i<n;i++)

         {

             cout<<savefro[i]<<' ';

         }

         cout<<endl;

         cout<<"saveb"<<endl;

         for(int i = 0;i<n;i++)

         {

             cout<<saveb[i]<<' ';

         }

         cout<<endl;

         cout<<"flag"<<endl;

         for(int i = 0;i<n;i++)

         {

             cout<<flag[i]<<' ';

         }

         cout<<endl;*/

         printf("%I64d\n",res);

     }

     return ;

 }