Army Creation

As you might remember from our previous rounds, Vova really likes computer games. Now he is playing a strategy game known as Rage of Empires.

In the game Vova can hire n different warriors; ith warrior has the type a_i. Vova wants to create a balanced army hiring some subset of warriors. An army is called balanced if for each type of warrior present in the game there are not more than k warriors of this type in the army. Of course, Vova wants his army to be as large as possible.

To make things more complicated, Vova has to consider q different plans of creating his army. ith plan allows him to hire only warriors whose numbers are not less than l_i and not greater than r_i.

Help Vova to determine the largest size of a balanced army for each plan.

Be aware that the plans are given in a modified way. See input section for details.

Input

The first line contains two integers n and k (1?≤?n,?k?≤?100000).

The second line contains n integers a₁, a₂, ... a_n (1?≤?a_i?≤?100000).

The third line contains one integer q (1?≤?q?≤?100000).

Then q lines follow. ith line contains two numbers x_i and y_i which represent ith plan (1?≤?x_i,?y_i?≤?n).

You have to keep track of the answer to the last plan (let‘s call it last). In the beginning last?=?0. Then to restore values of l_i and r_i for the ith plan, you have to do the following:

1. l_i?=?((x_i?+?last) mod n)?+?1;
2. r_i?=?((y_i?+?last) mod n)?+?1;
3. If l_i?>?r_i, swap l_i and r_i.

Output

Print q numbers. ith number must be equal to the maximum size of a balanced army when considering ith plan.

Example

Input

Copy

6 21 1 1 2 2 251 64 31 12 62 6

Output

Copy

24132

题解：b [ i ] 表示从 i 开始数 k 个和 a [ i ] 相同颜色的数的位置。那么在[ l , r ]区间内只有 b [ i ] 大于 r 的数才能被选中，等价于 [ l , r ] 区间内有多少个数大于r。套上主席树就能使用前缀和。算b[i]时要注意一下！！！

#pragma warning(disable:4996)
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define mem(arr,in) memset(arr,in,sizeof(arr))
using namespace std;

const int maxn = 1e5 + 5;

int n, m, q, cnt;
int root[maxn], a[maxn], b[maxn];

struct node { int l, r, sum; } T[maxn * 40];

vector<int> pos[maxn];

void Cal()
{
    for (int i = n; i >= 1; i--)
    {
        int sz = pos[a[i]].size();
        if (sz<m)
            b[i] = n + 1;
        else
            b[i] = pos[a[i]][sz - m];
        pos[a[i]].push_back(i);   //这句放在开头会WA！！！
    }
}

void Update(int l, int r, int &rt, int pre, int p) {
    T[++cnt] = T[pre], T[cnt].sum++, rt = cnt;
    if (l == r) return;
    int mid = (l + r) >> 1;
    if (mid >= p) Update(l, mid, T[rt].l, T[pre].l, p);
    else Update(mid + 1, r, T[rt].r, T[pre].r, p);
}

int Query(int l, int r, int rt, int L, int R) {
    if (l > R || r < L) return 0;
    if (L <= l && r <= R) return T[rt].sum;
    int mid = (l + r) >> 1;
    int ans = 0;
    ans += Query(l, mid, T[rt].l, L, R);
    ans += Query(mid + 1, r, T[rt].r, L, R);
    return ans;
}

int main()
{
    while (scanf("%d%d", &n, &m) != EOF) {
        cnt = 0;
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        Cal();
        for (int i = 1; i <= n; i++) Update(1, n + 1, root[i], root[i - 1], b[i]);
        scanf("%d", &q);
        int l, r, ans, last = 0;
        for (int i = 1; i <= q; i++) {
            scanf("%d%d", &l, &r);
            l = ((l + last) % n) + 1;
            r = ((r + last) % n) + 1;
            if (l > r) swap(l, r);
            ans = Query(1, n + 1, root[r], r + 1, n + 1) - Query(1, n + 1, root[l - 1], r + 1, n + 1);
            last = ans;
            printf("%d\n", ans);
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/zgglj-com/p/9053607.html