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package class_2023_03_1_week;
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// HH有一串由各种漂亮的贝壳组成的项链
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// HH 相信不同的贝壳会带来好运，所以每次散步完后，他都会随意取出一段贝壳，
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// 思考它们所表达的含义。HH 不断地收集新的贝壳，因此，他的项链变得越来越长。
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// 有一天，他突然提出了一个问题：某一段贝壳中，包含了多少种不同的贝壳？
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// 这个问题很难回答... 因为项链实在是太长了
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// 于是，他只好求助睿智的你，来解决这个问题
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// 测试链接 : https://www.luogu.com.cn/problem/P1972
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// 请同学们务必参考如下代码中关于输入、输出的处理
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// 这是输入输出处理效率很高的写法
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// 提交以下所有代码，把主类名改成Main
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// 洛谷对java太不友好了，大量时间不是消耗在算法本身上，而是耗在了IO上
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// 多提交几次能全通过
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import java.io.BufferedReader;
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import java.io.IOException;
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import java.io.InputStreamReader;
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import java.io.OutputStreamWriter;
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import java.io.PrintWriter;
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import java.io.StreamTokenizer;
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import java.util.Arrays;
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public class Code03_DiffColors {
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public static int MAXN = 1000010;
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public static int[] arr = new int[MAXN];
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public static int[][] query = new int[MAXN][3];
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public static int[] ans = new int[MAXN];
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public static int[] map = new int[MAXN];
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public static int[] tree = new int[MAXN];
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public static int n, m;
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public static void buildTree() {
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Arrays.fill(tree, 1, n + 1, 0);
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}
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public static int sum(int l, int r) {
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return sum(r) - sum(l - 1);
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}
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public static int sum(int index) {
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int ret = 0;
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while (index > 0) {
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ret += tree[index];
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index -= index & -index;
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}
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return ret;
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}
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public static void add(int i, int d) {
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while (i <= n) {
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tree[i] += d;
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i += i & -i;
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}
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}
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public static void main(String[] args) throws IOException {
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BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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StreamTokenizer in = new StreamTokenizer(br);
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PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
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while (in.nextToken() != StreamTokenizer.TT_EOF) {
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n = (int) in.nval;
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for (int i = 1; i <= n; i++) {
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in.nextToken();
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arr[i] = (int) in.nval;
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}
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// arr[i] : i位置的颜色
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in.nextToken();
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m = (int) in.nval;
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for (int i = 1; i <= m; i++) {
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in.nextToken();
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// left
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query[i][0] = (int) in.nval;
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in.nextToken();
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// right
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query[i][1] = (int) in.nval;
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// 第i个问题
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query[i][2] = i;
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}
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// map[5] = 8
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// 5这个颜色，上次出现在8位置
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// map[5] = 0
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// 5这个颜色，之前没出现过
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Arrays.fill(map, 0);
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buildTree();
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Arrays.sort(query, 1, m + 1, (a, b) -> a[1] - b[1]);
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for (int s = 1, j = 1; j <= m; j++) {
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int l = query[j][0];
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int r = query[j][1];
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int index = query[j][2];
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for (; s <= r; s++) {
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int color = arr[s];
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if (map[color] != 0) {
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add(map[color], -1);
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}
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add(s, 1);
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map[color] = s;
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}
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ans[index] = sum(l, r);
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}
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for (int i = 1; i <= m; i++) {
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out.println(ans[i]);
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}
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out.flush();
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}
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}
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}
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