[BZOJ2724]蒲公英

题意：Luogu
就是区间众数，强制在线
题解：
先离散化，分块大小\(O(\sqrt{\frac{n}{\log n}})\)(我也不知道怎么来的)
设\(f[i][j]\)表示第\(i,j\)块（包括两端点）的众数，预处理一下
对每个数开一个\(vector\)表示有这个数的位置，来计算任意两点间这个数的出现次数
区间众数显然只能是块间众数或者两遍剩余的块中的某个数
略为卡常
复杂度:\(O(n\sqrt{n})\)

#include<bits/stdc++.h>
using namespace std;
#define maxn 100005
inline void read(int& x)
{
	x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
}
int n, bsize;
int a[maxn], b[maxn], cnt[maxn], f[1005][1005], belong[maxn];
vector<int> ve[maxn];
inline void init()//块间预处理
{
	int mx, ans;
	register int i,j;
	for (i = 1; i <= belong[n]; ++i)
	{
		memset(cnt, 0, sizeof(cnt));
		mx = 0, ans = 0;
		for (j = (i - 1) * bsize + 1; j <= n; ++j)
		{
			++cnt[a[j]]; 
			if (cnt[a[j]] > mx || (cnt[a[j]] == mx && b[a[j]] < b[ans]))
				mx = cnt[a[j]], ans = a[j];
			f[i][belong[j]] = ans;
		}
	}
	for (i = 1; i <= n; ++i)//存位置
		ve[a[i]].push_back(i);
}
inline int query2(int l, int r, int k)//求k在[l,r]内出现了多少次
{
	return upper_bound(ve[k].begin(), ve[k].end(), r) - lower_bound(ve[k].begin(), ve[k].end(), l);
}
inline int query(int fr, int to)
{
	int ans, mx, tmp;
	register int i;
	ans = f[belong[fr] + 1][belong[to] - 1];//块间
	mx = query2(fr, to, ans);//块间众数出现次数
	for (i = fr; i <= min(to, belong[fr] * bsize); ++i)
	{
		tmp = query2(fr, to, a[i]);
		if (tmp > mx || (tmp == mx && b[a[i]] < b[ans]))
			mx = tmp, ans = a[i];
	}
	if (belong[fr] ^ belong[to])//if(belong[fr] != belong[to])
		for (i = (belong[to] - 1) * bsize + 1; i <= to; ++i)
		{
			tmp = query2(fr, to, a[i]);
			if (tmp > mx || (tmp == mx && b[a[i]] < b[ans]))
				mx = tmp, ans = a[i];
		}
	return ans;
}
int main()
{
	//freopen("testdata.in", "r", stdin),freopen("testout.out", "w", stdout);
	int y, z, m, x = 0;
	read(n), read(m);
	bsize = sqrt(n / log(n));
	register int i;
	for (i = 1; i <= ((n - 1) / bsize) + 1; ++i)
		for (int j = (i - 1) * bsize + 1; j <= i * bsize; ++j)
			belong[j] = i;
	for (i = 1; i <= n; ++i)
	{
		read(a[i]);
		b[i] = a[i];
	}
	sort(b + 1, b + n + 1);
	int k = unique(b + 1, b + n + 1) - b - 1;
	for (i = 1; i <= n; ++i)
		a[i] = lower_bound(b + 1, b + k + 1, a[i]) - b;
	init();
	for (i = 1; i <= m; ++i)
	{
		read(y); read(z);
		y = (y + x - 1) % n + 1, z = (z + x - 1) % n + 1;
		if (y > z) swap(y, z);
		x = b[query(y, z)];
		printf("%d\n", x);
	}
	return 0;
}