[HDU5919]Sequence II

[HDU5919]Sequence II

试题描述

Mr. Frog has an integer sequence of length n, which can be denoted as a₁,a₂,⋯,a_n There are m queries.

In the i-th query, you are given two integers li and ri. Consider the subsequence a_{l_i},a_{l_(i+1)},a_{l_(i+2)},⋯,a_ri.

We can denote the positions(the positions according to the original sequence) where an integer appears first in this subsequence as p⁽ⁱ⁾₁,p⁽ⁱ⁾₂,⋯,p⁽ⁱ⁾_{k_i} (in ascending order, i.e.,p⁽ⁱ⁾₁<p⁽ⁱ⁾₂<⋯<p⁽ⁱ⁾_{k_i}).

Note that k_i is the number of different integers in this subsequence. You should output p⁽ⁱ⁾_⌈ki/2⌉for the i-th query.

输入

In the first line of input, there is an integer T (T≤2) denoting the number of test cases.

Each test case starts with two integers n (n≤2×10⁵) and m (m≤2×10⁵). There are n integers in the next line, which indicate the integers in the sequence(i.e., a₁,a₂,⋯,a_n,0≤a_i≤2×10⁵).

There are two integers l_i and r_i in the following m lines.

However, Mr. Frog thought that this problem was too young too simple so he became angry. He modified each query to l‘_i,r‘_i(1≤l‘_i≤n,1≤r‘_i≤n). As a result, the problem became more exciting.

We can denote the answers as ans₁,ans₂,⋯,ans_m. Note that for each test case ans₀=0.

You can get the correct input l_i,r_i from what you read (we denote them as l‘_i,r‘_i)by the following formula:

l_i=min{(l‘_i+ans_i−1) mod n+1,(r‘_i+ans_i−1) mod n+1}

r_i=max{(l‘_i+ans_i−1) mod n+1,(r‘_i+ans_i−1) mod n+1}

输出

You should output one single line for each test case.

For each test case, output one line “Case #x: p₁,p₂,⋯,p_m”, where x is the case number (starting from 1) and p₁,p₂,⋯,p_m is the answer.

输入示例

2
5 2
3 3 1 5 4
2 2
4 4
5 2
2 5 2 1 2
2 3
2 4

输出示例

Case #1: 3 3
Case #2: 3 1

数据规模及约定

见“输入”

题解

就是这道题再强行套一个二分。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
	if(Head == Tail) {
		int l = fread(buffer, 1, BufferSize, stdin);
		Tail = (Head = buffer) + l;
	}
	return *Head++;
}
int read() {
	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 200010
#define maxnode 4000010

int ToT, sumv[maxnode], lc[maxnode], rc[maxnode];
void update(int& y, int x, int l, int r, int p) {
	sumv[y = ++ToT] = sumv[x] + 1;
	if(l == r) return ;
	int mid = l + r >> 1; lc[y] = lc[x]; rc[y] = rc[x];
	if(p <= mid) update(lc[y], lc[x], l, mid, p);
	else update(rc[y], rc[x], mid + 1, r, p);
	return ;
}
int query(int o, int l, int r, int qr) {
	if(!o) return 0;
	if(r <= qr) return sumv[o];
	int mid = l + r >> 1, ans = query(lc[o], l, mid, qr);
	if(qr > mid) ans += query(rc[o], mid + 1, r, qr);
	return ans;
}

int rt[maxn], lstp[maxn], ANS[maxn], cnt;

int len;
char Out[maxn*7];
int main() {
	int T = read();
	for(int kase = 1; kase <= T; kase++) {
		memset(lstp, 0, sizeof(lstp));
		memset(sumv, 0, sizeof(sumv));
		memset(lc, 0, sizeof(lc));
		memset(rc, 0, sizeof(rc));
		memset(rt, 0, sizeof(rt));
		ToT = 0;
		int n = read(), q = read();
		for(int i = 1; i <= n; i++) {
			int v = read();
			update(rt[i], rt[i-1], 0, n, lstp[v]);
			lstp[v] = i;
		}
		
		cnt = 0;
		int lst = 0;
		while(q--) {
			int ql = (read() + lst) % n + 1, qr = (read() + lst) % n + 1;
			if(ql > qr) swap(ql, qr);
			int l = ql, r = qr, k = query(rt[qr], 0, n, ql - 1) - query(rt[ql-1], 0, n, ql - 1) + 1 >> 1, lval = query(rt[ql-1], 0, n, ql - 1);
			while(l < r) {
				int mid = l + r >> 1;
				if(query(rt[mid], 0, n, ql - 1) - lval < k)
					l = mid + 1;
				else r = mid;
			}
			ANS[++cnt] = lst = l;
		}
		
		printf("Case #%d: ", kase);
		len = 0;
		int num[10], cntn;
		for(int i = 1; i <= cnt; i++) {
			int tmp = ANS[i];
			if(!tmp) Out[len++] = '0';
			cntn = 0; while(tmp) num[++cntn] = tmp % 10, tmp /= 10;
			for(int j = cntn; j; j--) Out[len++] = num[j] + '0';
			if(i < cnt) Out[len++] = ' ';
		}
		Out[len] = '\0';
		puts(Out);
	}
	
	return 0;
}