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Given a sequence of n numbers a1, a2, ..., an and a number of d-queries. A d-query is a pair (i, j) (1 ≤ i ≤ j ≤ n). For each d-query (i, j), you have to return the number of distinct elements in the subsequence ai, ai+1, ..., aj.
Input
- Line 1: n (1 ≤ n ≤ 30000).
- Line 2: n numbers a1, a2, ..., an (1 ≤ ai ≤ 106).
- Line 3: q (1 ≤ q ≤ 200000), the number of d-queries.
- In the next q lines, each line contains 2 numbers i, j representing a d-query (1 ≤ i ≤ j ≤ n).
Output
- For each d-query (i, j), print the number of distinct elements in the subsequence ai, ai+1, ..., aj in a single line.
Example
Input 5 1 1 2 1 3 3 1 5 2 4 3 5 Output 3 2 3
查找给定区间内有多少不同的数。
以每个数最后出现的坐标为点建立主席树。这样的话对于【L,R】区间内的不同的数的数量就是主席树【R】中大于等于L的点的数量是多少。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 30010;
const int M = maxn * 200;
int n, q, tot;
int a[maxn];
int T[maxn], lson[maxn], rson[maxn], c[M];
int build(int l, int r) {
int root = tot++;
c[root] = 0;
if (l != r) {
int mid = (l + r) >> 1;
lson[root] = build(l, mid);
rson[root] = build(mid + 1, r);
}
return root;
}
int update(int root, int pos, int val) {
int newroot = tot++, tmp = newroot;
c[newroot] = c[root] + val;
int l = 1, r = n;
while (l < r) {
int mid = (l + r) >> 1;
if (pos <= mid) {
lson[newroot] = tot++; rson[newroot] = rson[root];
newroot = lson[newroot]; root = lson[root];
r = mid;
}
else {
rson[newroot] = tot++; lson[newroot] = lson[root];
newroot = rson[newroot]; root = rson[root];
l = mid + 1;
}
c[newroot] = c[root] + val;
}
return tmp;
}
int query(int l, int r, int rt, int L) {
if (l >= L) {
return c[rt];
}
int mid = (l + r) >> 1, sum = 0;
if (L <= mid) {
sum += query(l, mid, lson[rt], L);
sum += c[rson[rt]];
}
else sum += query(mid + 1, r, rson[rt], L);
return sum;
}
int main() {
while (~scanf("%d", &n)) {
tot = 0;
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
T[0] = build(1, n);
map<int, int>G;
for (int i = 1; i <= n; i++) {
if (G.find(a[i]) == G.end()) {
T[i] = update(T[i - 1], i, 1);
}
else {
int tmp = update(T[i - 1], G[a[i]], -1);
T[i] = update(tmp, i, 1);
}
G[a[i]] = i;
}
scanf("%d", &q);
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n", query(1, n, T[r], l));
}
}
return 0;
}