题意

\(n\)个数\(a[i] ,q\)次询问,\(n,a[i],q<=10^5\)每次问\([l,r]\)内最多可以选多少个数,满足同一个数的出现次数不超过\(k\)

强制在线

Sol

处理出每个数往前数第\(k+1\)个与它相同的位置

没有则为\(0\)

那么就是求区间内所有的该值小于\(l\)的数

主席树来做就好了

# include <bits/stdc++.h>

# define RG register

# define IL inline

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

const int _(1e5 + 5);

IL int Input(){

    RG int x = 0, z = 1; RG char c = getchar();

    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;

    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);

    return x * z;

}

int n, k, rt[_], tot;

struct HJT{

	int ls, rs, sz;

} T[_ * 20];

vector <int> num[_];

IL void Modify(RG int &x, RG int l, RG int r, RG int p){

	T[++tot] = T[x], ++T[x = tot].sz;

	if(l == r) return;

	RG int mid = (l + r) >> 1;

	if(p <= mid) Modify(T[x].ls, l, mid, p);

	else Modify(T[x].rs, mid + 1, r, p);

}

IL int Query(RG int x, RG int l, RG int r, RG int p){

	if(!x) return 0;

	if(l == r) return T[x].sz;

	RG int mid = (l + r) >> 1;

	if(p <= mid) return Query(T[x].ls, l, mid, p);

	return T[T[x].ls].sz + Query(T[x].rs, mid + 1, r, p);

}

int main(RG int argc, RG char* argv[]){

	n = Input(), k = Input();

	for(RG int i = 1, a, b; i <= n; ++i){

		a = Input(), num[a].push_back(i);

		RG int l = num[a].size();

		if(l <= k) b = 0;

		else b = num[a][l - k - 1];

		rt[i] = rt[i - 1], Modify(rt[i], 0, n, b);

	}

	for(RG int q = Input(), ans = 0; q; --q){

		RG int l = Input(), r = Input();

		l = (l + ans) % n + 1, r = (r + ans) % n + 1;

		if(l > r) swap(l, r);

		printf("%d\n", ans = Query(rt[r], 0, n, l - 1) - Query(rt[l - 1], 0, n, l - 1));

	}

    return 0;

}