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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> 樹狀數組Codeforces Round #261 (Div. 2)D

樹狀數組Codeforces Round #261 (Div. 2)D

編輯:C++入門知識

樹狀數組Codeforces Round #261 (Div. 2)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 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


和求逆序對差不多,需要先預處理一下


#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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