[HDU5919]Sequence II

试题描述

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

In the i-th query, you are given two integers li and ri. Consider the subsequence al_i,al_(i+1),al_(i+2),⋯,ari.

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 ki 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,⋯,an,0≤ai≤2×105).

There are two integers li and ri 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,⋯,ansm. Note that for each test case ans0=0.

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

li=min{(l‘i+ansi−1) mod n+1,(r‘i+ansi−1) mod n+1}
ri=max{(l‘i+ansi−1) mod n+1,(r‘i+ansi−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,⋯,pm”, where x is the case number (starting from 1) and p1,p2,⋯,pm is the answer.

输入示例


输出示例

Case #:
Case #:

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;
}

[HDU5919]Sequence II的更多相关文章

  1. HDU5919 Sequence II(主席树)

    Mr. Frog has an integer sequence of length n, which can be denoted as a1,a2,⋯,ana1,a2,⋯,anThere are ...

  2. HDU 5919 Sequence II 主席树

    Sequence II Problem Description   Mr. Frog has an integer sequence of length n, which can be denoted ...

  3. HDU 5919 Sequence II(主席树+逆序思想)

    Sequence II Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) To ...

  4. HDOJ 5147 Sequence II 树阵

    树阵: 每个号码的前面维修比其数数少,和大量的这后一种数比他的数字 再枚举每一个位置组合一下 Sequence II Time Limit: 5000/2500 MS (Java/Others)    ...

  5. bestcoder#23 1002 Sequence II 树状数组+DP

    Sequence II Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total ...

  6. hdu 5147 Sequence II 树状数组

    Sequence II Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Prob ...

  7. hdu 5147 Sequence II (树状数组 求逆序数)

    题目链接 Sequence II Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  8. Sequence II

    6990: Sequence II 时间限制: 3 Sec  内存限制: 128 MB提交: 206  解决: 23[提交][状态][讨论版][命题人:admin] 题目描述 We define an ...

  9. hdu 5147 Sequence II【树状数组/线段树】

    Sequence IITime Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem ...

随机推荐

  1. LeetCode 69 题

    1.题目要求 实现 int sqrt(int x) 函数. 计算并返回 x 的平方根,其中 x 是非负整数. 由于返回类型是整数,结果只保留整数的部分,小数部分将被舍去. 示例 1: 输入: 4 输出 ...

  2. 基于socketserver实现的并发(tcp和udp)

    threading 线程 基于tcp协议:请求建立连接,然后开启进程 基于udp协议:直接开启新进程 基于tcp协议 import socketserver # 导入socketserver模块 # ...

  3. IO 优化

    转自 BlackJack_ #define fastcall __attribute__((optimize("-O3"))) #define IL __inline__ __at ...

  4. php 缓存工具类 实现网页缓存

    php 缓存工具类 实现网页缓存 php程序在抵抗大流量访问的时候动态网站往往都是难以招架,所以要引入缓存机制,一般情况下有两种类型缓存 一.文件缓存 二.数据查询结果缓存,使用内存来实现高速缓存 本 ...

  5. 数据源引用 java:/comp/env

      编辑 删除 数据源引用 java:/comp/env 2012-01-28 15:59   ENC的概念: The application component environment is ref ...

  6. UVA 10288 Coupons 彩票 (数学期望)

    题意:一种刮刮卡一共有n种图案,每张可刮出一个图案,收集n种就有奖,问平均情况下买多少张才能中奖?用最简的分数形式表示答案.n<=33. 思路:这题实在好人,n<=33.用longlong ...

  7. php bz2扩展安装

    php bz2扩展安装 2017年09月22日 14:14:36 Cookie_1030 阅读数:1781   版权声明:本文为博主原创文章,未经博主允许不得转载. https://blog.csdn ...

  8. 看云&gitbook 写帮助文档 | 专注于文档在线创作、协作和托管

    看云 写帮助文档 | 专注于文档在线创作.协作和托管 https://www.kancloud.cn/manual/thinkphp/1678 https://www.gitbook.com/

  9. C ++ _基础之共用体

    由以下代码来进一步学习共用体 #include <stdio.h> #include<iostream> void main() { union un { int a; cha ...

  10. Redis进阶例子

    工作中用到的RabbitMQ例子 , 但是最后没有用 , 用的CMQ , 顺便说下CMQ社区真的少 , 并且功能少 . 一.消息体 package com.bootdo.common.rabbitmq ...