Codeforces 368B Sereja and Suffixes 线段树按值建树_10 10 1 2 3 4 1 2 3 4 100000 99999 1 2 3 4 5 6 7 8-优快云博客

本文介绍了一个经典的算法问题——Sereja与后缀问题，该问题要求在给定的整数数组中，对于一系列指定的起始位置，找出从该位置到数组末尾的不同元素数量。文中提供了一种使用线段树进行预处理的方法，以及一种更高效的DP解决方案。

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B. Sereja and Suffixes

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Sereja has an array a, consisting of n integers a₁, a₂, ..., a_n. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out m integers l₁, l₂, ..., l_m (1 ≤ l_i ≤ n). For each number l_i he wants to know how many distinct numbers are staying on the positions l_i, l_i + 1, ..., n. Formally, he want to find the number of distinct numbers among a_{l_i}, a_{l_i + 1}, ..., a_n.?

Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each l_i.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 10⁵). The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁵) — the array elements.

Next m lines contain integers l₁, l₂, ..., l_m. The i-th line contains integer l_i (1 ≤ l_i ≤ n).

Output

Print m lines — on the i-th line print the answer to the number l_i.

Sample test(s)

Input

10 10
1 2 3 4 1 2 3 4 100000 99999
1
2
3
4
5
6
7
8
9
10

Output

题意：给出一个数列长度为n，m次询问，每次输入一个l，要求输出区间[l,n]中有多少个不同的数。

思路：用线段数进行预处理，处理出每个位置到n的这个区间内不同数的个数。将线段树按值建树，从数列的尾部往前面扫描，对于数列的每一个数a[i],在线段树中，如果a[i]处为0的话，把它改成1，否则不变，然后再对整个线段树的区间求和，就是第i个位置的答案。

代码：

#include<cstdio>
#include<cstring>
#define maxn 100005
int sum[maxn << 2];
void update(int id, int L, int R, int pos)
{
	if (L == R) sum[id] = 1;
	else
	{
		int mid = (L + R) >> 1;
		if (pos <= mid) update(id << 1, L, mid, pos);
		else update(id << 1 | 1, mid + 1, R, pos);
		sum[id] = sum[id << 1] + sum[id << 1 | 1];
	}
}
int a[maxn];
int ans[maxn];
int main()
{
	int n, m;
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	memset(sum, 0, sizeof(sum));
	for (int i = n; i >= 1; i--)
	{
		update(1, 1, 100000, a[i]);
		ans[i] = sum[1];
	}
	while (m--)
	{
		int x;
		scanf("%d", &x);
		printf("%d\n", ans[x]);
	}
	return 0;
}

还有就是DP的做法，时间复杂度仅为O(n)，dp[i]=dp[i+1]+(vis[ai]]);

用bool vis[i]记录是否出现过。