CF #261 Div2 D. Pashmak and Parmida's problem （离散化+逆序对+线段树） - c++编程基础

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,?...,?a_n. Let's denote f(l,?r,?x) the number of indices k such that: l?≤?k?≤?r and a_k?=?x. His task is to calculate the number of pairs of indicies i,?j (1?≤?i? j?≤?n) such that f(1,?i,?a_i)?>?f(j,?n,?a_j).

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,?...,?a_n (1?≤?a_i?≤?10⁹).

Output

Print a single integer ― the answer to the problem.

Sample test(s) Input

7
1 2 1 1 2 2 1

Output

Input

3
1 1 1

Output

Input

5
1 2 3 4 5

Output

题目给出的F函数，可以用离散的方法加预处理将每个F（1，i，x）和F（j，n，x）求出，分别保存于f1, f2数组，那么题目就可以转化为： f1[i] > f2[j] && i < j , 求出满足条件的个数，和逆序对的思想基本一样，但本题不好用归并排序解决，因为牵扯到两个数组，那么可以考虑线段树解决。

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         #define lson o << 1, l, m #define rson o << 1|1, m+1, r using namespace std; typedef long long LL; const int MAX = 200000000; const int maxn = 1000100; int n, a, b; int vis[maxn], tt[maxn], in[maxn], f1[maxn], f2[maxn], num[maxn<<2], fu[maxn]; int bs(int v, int x, int y) { int m; while(x < y) { m = (x+y) >> 1; if(fu[m] >= v) y = m; else x = m+1; } return x; } void up(int o) { num[o] = num[o<<1] + num[o<<1|1]; } void update(int o, int l, int r) { if(l == r) num[o]++; else { int m = (l+r) >> 1; if(a <= m) update(lson); else update(rson); up(o); } } LL query(int o, int l, int r) { if(a <= l && r <= b) return num[o]; int m = (l+r) >> 1; LL res = 0; if(a <= m) res += query(lson); if(m < b ) res += query(rson); return res; } int main() { cin >> n; for(int i = 0; i < n; i++) { scanf("%d", &in[i]); tt[i] = in[i]; } sort(in, in+n); int k = 0; fu[k++] = in[0]; for(int i = 1; i < n; i++) if(in[i] != in[i-1]) fu[k++] = in[i]; for(int i = 0; i < n; i++) { int tmp = bs(tt[i], 0, k-1); vis[tmp]++; f1[i] = vis[tmp]; } memset(vis, 0, sizeof(vis)); for(int i = n-1; i >= 0; i--) { int tmp = bs(tt[i], 0, k-1); vis[tmp]++; f2[i] = vis[tmp]; b = max(b, f2[i]); } LL ans = 0; for(int i = 0; i < n; i++) { a = f2[i]+1; ans += query(1, 0, b); a = f1[i]; update(1, 0, b); } cout << ans << endl; return 0; }