SPOJ DQUERY 求区间内不同数的个数 （主席树）

DQUERY - D-query

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

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 2000010;

struct node {
int l, r, ls, rs, cont;
}t
;
int n, q, a
, lastans, tot = 0, num, tail = 0, prev
, temp
, root
;

void update(int root) {
t[root].cont = t[t[root].ls].cont + t[t[root].rs].cont;
}

int build(int l, int r) {
int num = ++ tail;
t[num].l = l, t[num].r = r;
if(l == r)
return num;
int mid = (l + r) >> 1;
t[num].ls = build(l, mid);
t[num].rs = build(mid + 1, r);
return num;
}

int modify(int pre, int pos) {
int num = ++ tail;
t[num] = t[pre];
if(t[num].l == t[num].r) {
t[num].cont ++;
return num;
}
int mid = (t[num].l + t[num].r) >> 1;
if(pos <= mid)
t[num].ls = modify(t[num].ls, pos);
else
t[num].rs = modify(t[num].rs, pos);
update(num);
return num;
}

int query(int num1, int num2, int l, int r) {
if(t[num1].l == l && t[num1].r == r)
return t[num2].cont - t[num1].cont;
int mid = (t[num1].l + t[num1].r) >> 1;
if(r <= mid)
return query(t[num1].ls, t[num2].ls, l, r);
else if(l > mid)
return query(t[num1].rs, t[num2].rs, l, r);
else
return query(t[num1].ls, t[num2].ls, l, mid) + query(t[num1].rs, t[num2].rs, mid + 1, r);
}

int main() {
scanf("%d", &n);
for(register int i = 1; i <= n; i ++) {
scanf("%d", &a[i]);
temp[i] = a[i];
}
sort(temp + 1, temp + n + 1);
tot = unique(temp + 1, temp + n + 1) - temp - 1;
for(int i = 1; i <= n; i ++)
a[i] = lower_bound(temp + 1, temp + tot + 1, a[i]) - temp;
root[0] = build(0, n);
for(int i = 1; i <= n; i ++) {
root[i] = modify(root[i - 1], prev[a[i]]);
prev[a[i]] = i;
}
scanf("%d", &q);
while(q --) {
int l, r;
scanf("%d %d", &l, &r);
lastans = r - l + 1 - query(root[l - 1], root[r], l, r);
printf("%d\n", lastans);
}
return 0;
}

上面是我不知道能不能A的代码