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树状数组Codeforces Round #261 (Div. 2)D
2015-07-20 17:53:41 来源: 作者: 【 】 浏览:2
Tags:Codeforces Round #261 Div.

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 and ak?=?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.

Sample test(s) input
7
1 2 1 1 2 2 1
output
8
input
3
1 1 1
output
1
input
5
1 2 3 4 5
output
0


和求逆序对差不多,需要先预处理一下


#include
  
   
#include
   
     #include
    
      #include
      #include
      
        #include
       
         #include
        
          using namespace std; typedef long long LL; #define maxn (int)1e6+5 int A[maxn],c[maxn],b[maxn],s[maxn]; int n; void add(int x,int d) { while(x<=n) { s[x]+=d; x+=(x&-x); } } LL Query(int x) { LL sum=0; while(x>0) { sum+=s[x]; x-=(x&-x); } return sum; } typedef long long LL; int main() { while(~scanf("%d",&n)) { map
         
           q; for(int i=1;i<=n;i++) { scanf("%d",A+i); q[A[i]]++; b[i]=q[A[i]]; } q.clear(); memset(s,0,sizeof(s)); for(int i=n;i>=2;i--) { q[A[i]]++; c[i]=q[A[i]]; add(c[i],1); } LL ans=0; for(int i=1;i
          
           


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