[BZOJ2724][Violet 6] The dandelion

Test description Input Fix it

l = (l_0 + x - 1) mod n + 1, r = (r_0 + x - 1) mod n + 1

Output Input example

Output example

Data scale and agreement Fix it ：

n <= 40000, m <= 50000

Block , Preprocess first f[i][j] It means the first one i Block to j The mode of the block , Enumeration starting point i And then just sweep it .

The second is to ask , For an inquiry [ql, qr], among ql Belong to block l,qr Belong to block r, The mode is either f[l+1][r-1], Or numbers in incomplete blocks , So we need to build a calc(l, r, x) function , Asking [l, r] in x Number of occurrences , With this function, you can initially set the answer to f[l+1][r-1]（O(1)）, then O(sqrt(n)) Brute force enumeration of numbers in incomplete blocks , If it happens more than the current one , Just update the answer .

This calc() Functions can do this ： Discrete all the numbers , Write down the position of each number , And then when you ask calc(l, r, x) stay x It can be calculated by bisecting the position sequence of .

```#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
const int BufferSize = 1 << 16;
inline char Getchar() {
int l = fread(buffer, 1, BufferSize, stdin);
Tail = (Head = buffer) + l;
}
}
int x = 0, f = 1; char c = Getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = Getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = Getchar(); }
return x * f;
}
#define maxn 40010
#define maxq 210
int n, m, num[maxn], A[maxn], st[maxq], en[maxq], bl[maxn], cntb, f[maxq][maxq], cntn[maxn];
vector <int> pos[maxn];
int calc(int l, int r, int x) {
return upper_bound(pos[x].begin(), pos[x].end(), r) - lower_bound(pos[x].begin(), pos[x].end(), l);
}
int query(int ql, int qr) {
int l = bl[ql], r = bl[qr];
if(r - l <= 1) {
int ans = A[ql], mx = calc(ql, qr, ans);
for(int i = ql + 1; i <= qr; i++) {
int tmp = calc(ql, qr, A[i]);
if(mx < tmp || (mx == tmp && ans > A[i])) ans = A[i], mx = tmp;
}
return ans;
}
int ans = f[l+1][r-1], mx = calc(ql, qr, ans);
for(int i = ql; i <= en[l]; i++) {
int tmp = calc(ql, qr, A[i]);
if(mx < tmp || (mx == tmp && ans > A[i])) ans = A[i], mx = tmp;
}
for(int i = st[r]; i <= qr; i++) {
int tmp = calc(ql, qr, A[i]);
if(mx < tmp || (mx == tmp && ans > A[i])) ans = A[i], mx = tmp;
}
return ans;
}
int main() {
int siz = (int)sqrt(n + .5);
for(int i = 1; i <= n; i++) {
int p = (i - 1) / siz + 1; cntb = p;
if(!st[p]) st[p] = i;
en[p] = i;
bl[i] = p;
}
sort(num + 1, num + n + 1);
for(int i = 1; i <= n; i++) A[i] = lower_bound(num + 1, num + n + 1, A[i]) - num;
for(int i = 1; i <= cntb; i++) {
memset(cntn, 0, sizeof(cntn));
for(int j = st[i]; j <= n; j++) {
cntn[A[j]]++;
int r = bl[j];
if(r > i && !f[i][r]) f[i][r] = f[i][r-1];
if(!f[i][r] || cntn[f[i][r]] < cntn[A[j]] || (cntn[f[i][r]] == cntn[A[j]] && f[i][r] > A[j]))
f[i][r] = A[j];
}
}
for(int i = 1; i <= n; i++) pos[A[i]].push_back(i);
int x = 0;
while(m--) {
int l = (read() + x - 1) % n + 1, r = (read() + x - 1) % n + 1;
if(l > r) swap(l, r);
int tmp = query(l, r);
printf("%d\n", num[tmp]);
x = num[tmp];
}
return 0;
}
```

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