Pashmak and Parmida's problem（树状数组）

题目链接：http://codeforces.com/contest/459/problem/D

题意： 数列A， ai表示 i-th 的值， f（i，j, x） 表示【i，j】之间x的数目， 问：当 1 <= i < j <= n， 满足 f(1, i, ai) < f（j，n, aj）的对数。

题解：分别求出两组数列  Ii = f(1, i, ai)（1<=i <= n） Jj =  f（j，n, aj）(1 <= j <= n), 然后， 从n开始， 在数列   Ij（i < j <= n）中求出   小于 Ii的个数和。

 /***Good Luck***/
 #define _CRT_SECURE_NO_WARNINGS
 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <stack>
 #include <map>
 #include <queue>
 #include <vector>
 #include <set>
 #include <functional>
 #include <cmath>

 #define Zero(a) memset(a, 0, sizeof(a))
 #define Neg(a)  memset(a, -1, sizeof(a))
 #define All(a) a.begin(), a.end()
 #define PB push_back
 #define inf 0x3f3f3f3f
 #define inf2 0x7fffffffffffffff
 #define ll long long
 using namespace std;
 //#pragma comment(linker, "/STACK:102400000,102400000")
 void get_val(int &a) {
     int value = , s = ;
     char c;
     while ((c = getchar()) == ' ' || c == '\n');
     if (c == '-') s = -s; else value = c - ;
     while ((c = getchar()) >= '' && c <= '')
         value = value *  + c - ;
     a = s * value;
 }
 const int maxn = ;
 int arri[maxn], arrj[maxn];
 int arr2[maxn];
 int arr1[maxn];
 int A[maxn];
 int n;
 int  lowbit(int x) {
     return x & (-x);
 }

 int sum(int i) {
     int ret = ;
     while (i > ) {
         ret += A[i];
         i -= lowbit(i);
     }
     return ret;
 }

 void update(int pos, int val) {
     while (pos <= n) {
         A[pos] += val;
         pos += lowbit(pos);
     }
 }

 int main() {
     //freopen("data.out", "w", stdout);
     //freopen("data.in", "r", stdin);
     //cin.sync_with_stdio(false);

     scanf("%d", &n);
     for (int i = ; i <= n; ++i) {
         scanf("%d", arr1 + i);
         arr2[i] = arr1[i];
     }

     sort(arr2, arr2 + n + );           // 离散化
     int len = unique(arr2, arr2 + n) - arr2;
     for (int i = ; i <= n; ++i) {
         arr1[i] = lower_bound(arr2, arr2 + len, arr1[i]) - arr2;
     }
     Zero(arr2);
     for (int i = ; i <= n; ++i) {
         arri[i] = ++arr2[arr1[i]];

     }
     Zero(arr2);
     for (int i = n ; i > ; --i) {
         arrj[i] = ++arr2[arr1[i]];
     }
     ll count = ;
     for (int i = n; i >= ; --i) {

         count += sum(arri[i] - ) ;
         update(arrj[i], );
     }
     cout << count << endl;
     return ;
 }