本文主要是介绍CodeForces 459D Pashmak and Parmida's problem，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目链接：http://codeforces.com/problemset/problem/459/D
Pashmak and Parmida's problem
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.
Examples
7
1 2 1 1 2 2 1
8
3
1 1 1
1
5
1 2 3 4 5
0
思路：先正着、逆着分别将各个区间的值算出来，然后对于逆着的区间，只有小于其下标的值才会影响答案，因此，对于每个正着的区间，在计算其结果时，大于其下标的必须要被判断计算，这样就很容易想到用树状数组来维护，将正着的区间倒着计算，并将大于其下标的值插入树状数组计数即可。注意：本题数据可能很大，开始计数可以选择用map来实现。详见代码。
附上AC代码：
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;
typedef long long ll;
const int maxn = 1000005;
int num[maxn], a[maxn], b[maxn];
int n;
map<int, int> cnt;int lowbit(int x){return x&(-x);
}void add(int p){while (p <= n){++num[p];p += lowbit(p);}
}ll sum(int p){ll ans = 0;while (p > 0){ans += num[p];p -= lowbit(p);}return ans;
}int main(){while (~scanf("%d", &n)){for (int i=1; i<=n; ++i)scanf("%d", num+i);cnt.clear();for (int i=1; i<=n; ++i){++cnt[num[i]];a[i] = cnt[num[i]];}cnt.clear();for (int i=n; i>=1; --i){++cnt[num[i]];b[i] = cnt[num[i]];}
// for (int i=1; i<=n; ++i)
// printf("%d%c", a[i], i!=n ? ' ' : '\n');
// for (int i=1; i<=n; ++i)
// printf("%d%c", b[i], i!=n ? ' ' : '\n');memset(num, 0, sizeof(num));ll ans = 0;for (int i=n-1; i>=1; --i){add(b[i+1]);//printf("%I64d\n", sum(a[i]-1));ans += sum(a[i]-1);}printf("%I64d\n", ans);}return 0;
}
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