Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P4168 [Violet]蒲公英

P4168 [Violet]蒲公英

区间众数的板子。

确定完一个区间所有的数出现的个数后,一个瓶颈是枚举哪些数可能成为众数。可以发现众数可能出现的位置是中间整块的众数和边角的数,这样将其降为 $O(\sqrt n)$ ,另外查询区间块的众数,需要前缀和方便 $O(1)$ 查询。

查看代码
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#include <cstdio>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
(c == '-') && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
(flag) && (x = ~x + 1);
}
const int N = 4e4 + 10, M = 210;
int n, m, len, w[N], s[M][N], p[M][M], cnt[N];
int query(int l, int r)
{
if (r / len - l / len < 2)
{
int t = -1;
for (int i = l; i <= r; i++)
{
cnt[w[i]]++;
if (t == -1 || cnt[w[i]] > cnt[t] || (cnt[w[i]] == cnt[t] && w[i] < t))
t = w[i];
}
for (int i = l; i <= r; i++)
cnt[w[i]]--;
return t;
}
int t = p[l / len + 1][r / len - 1], mx = s[r / len - 1][t] - s[l / len][t];
for (int i = l; i < l / len * len + len; i++)
{
cnt[w[i]]++;
int v = cnt[w[i]] + s[r / len - 1][w[i]] - s[l / len][w[i]];
if (v > mx || (v == mx && w[i] < t))
{
t = w[i];
mx = v;
}
}
for (int i = r / len * len; i <= r; i++)
{
cnt[w[i]]++;
int v = cnt[w[i]] + s[r / len - 1][w[i]] - s[l / len][w[i]];
if (v > mx || (v == mx && w[i] < t))
{
t = w[i];
mx = v;
}
}
for (int i = l; i < l / len * len + len; i++)
cnt[w[i]]--;
for (int i = r / len * len; i <= r; i++)
cnt[w[i]]--;
return t;
}
int main()
{
read(n), read(m);
len = sqrt(n);
vector<int> ws;
for (int i = 0; i < n; i++)
{
read(w[i]);
ws.push_back(w[i]);
}
sort(ws.begin(), ws.end());
ws.erase(unique(ws.begin(), ws.end()), ws.end());
for (int i = 0; i < n; i++)
w[i] = lower_bound(ws.begin(), ws.end(), w[i]) - ws.begin();
for (int i = 0; i < n; i++)
s[i / len][w[i]]++;
for (int i = 1; i <= (n - 1) / len; i++)
for (int j = 0; j < n; j++)
s[i][j] += s[i - 1][j];
for (int i = 0; i <= (n - 1) / len; i++)
{
for (int j = i, t = -1; j <= (n - 1) / len; j++)
{
for (int k = j * len; k < min(n, j * len + len); k++)
{
cnt[w[k]]++;
if (t == -1 || cnt[w[k]] > cnt[t] || (cnt[w[k]] == cnt[t] && w[k] < t))
t = w[k];
}
p[i][j] = t;
}
for (int j = 0; j < n; j++)
cnt[j] = 0;
}
for (int l, r, last = 0; m; m--)
{
read(l), read(r);
l = (l + last - 1) % n, r = (r + last - 1) % n;
if (l > r)
swap(l, r);
printf("%d\n", last = ws[query(l, r)]);
}
return 0;
}