题意

询问区间有多少不同的数。

题解

和Luogu 1903一样，这道题也是用pre数组来求区间不同数的个数，这里pre[i]表示a[i]上一次出现的位置 +１，询问相当于查询区间内有多少pre小于等于左端点。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
#define space putchar(' ')
#define enter putchar('\n')
template <class T>
void read(T &x){char c;bool op = 0;while(c = getchar(), c < '0' || c > '9')if(c == '-') op = 1;x = c - '0';while(c = getchar(), c >= '0' && c <= '9')x = x * 10 + c - '0';if(op) x = -x;
}
template <class T>
void write(T x){if(x < 0) x = -x, putchar('-');if(x >= 10) write(x / 10);putchar('0' + x % 10);
}
const int N = 50005;
int n, m, l, r, ans, a[N], last[1000005], pre[N];
int idx, ls[N*50], rs[N*50], data[N*50], root[N];
void build(int &k, int l, int r){k = ++idx;if(l == r) return;int mid = (l + r) >> 1;build(ls[k], l, mid);build(rs[k], mid + 1, r);
}
void change(int old, int &k, int l, int r, int p, int x){k = ++idx;ls[k] = ls[old], rs[k] = rs[old], data[k] = data[old];if(l == r) return (void)(data[k] += x);int mid = (l + r) >> 1;if(p <= mid) change(ls[old], ls[k], l, mid, p, x);else change(rs[old], rs[k], mid + 1, r, p, x);data[k] = data[ls[k]] + data[rs[k]];
}
int query(int old, int k, int l, int r, int ql, int qr){if(ql <= l && qr >= r) return data[k] - data[old];int mid = (l + r) >> 1, ret = 0;if(ql <= mid) ret += query(ls[old], ls[k], l, mid, ql, qr);if(qr > mid) ret += query(rs[old], rs[k], mid + 1, r, ql, qr);return ret;
}
int main(){read(n);build(root[0], 1, n);for(int i = 1; i <= n; i++){read(a[i]);pre[i] = last[a[i]] + 1;last[a[i]] = i;change(root[i - 1], root[i], 1, n, pre[i], 1);}read(m);while(m--){read(l), read(r);write(query(root[l - 1], root[r], 1, n, 1, l)), enter;}return 0;
}