codeforces Round #261(div2) DProblem Solution Report_html/css

codeforces Round #261(div2) DProblem Solution Report_html/css_WEB-ITnose

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Release： 2016-06-24 11:55:51

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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??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

71 2 1 1 2 2 1

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output

input

31 1 1

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output

input

51 2 3 4 5

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output

题目大意：

给出数组A，定义f(l,r,x)为A[]的下标l到r之间，等于x的元素数。i和j符合f(1,i,a[i])>f(j,n,a[j])，求有多少对这样的(i,j).

解法：

我们可以通过预处理，在O(n)之内，算出f(1, i, a[i]) 定义数组为l[i]，算出f(j, n, a[j]) 定义数组为r[i]。

题目就转化为：求出 l[i] > r[j] (i < j)。

本质上，就是一个求逆序对的做法，用树状数组。

代码：

#include #include #include #define LL long long#define N_max 3*1234567using namespace std;map  l, r;int n;int bit[N_max], a[N_max];LL ans;void init() {	scanf("%d", &n);	for (int i = 1; i <= n; i++)  scanf("%d", &a[i]);}void add(int i, int x) {	while (i <= n) {		bit[i] += x;		i += i&-i;	}}LL sum(int i) {	LL s = 0;	while (i > 0) {		s += bit[i];		i -= i&-i;	}	return s;}void solve() {	l.clear();	r.clear();	for (int i = n; i >= 1; i--) {		l[a[i]]++;		add(l[a[i]], 1);	}	for (int i = 1; i <= n; i++) {		r[a[i]]++;		add(l[a[i]], -1);		l[a[i]]--;				ans += sum(r[a[i]]-1);	}	cout << ans << endl;}int main() {	init();	solve();}