[HDU5919]Sequence II
[HDU5919]Sequence II
试题描述
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
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:
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
- 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 ...
- HDU 5919 Sequence II 主席树
Sequence II Problem Description Mr. Frog has an integer sequence of length n, which can be denoted ...
- 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 ...
- 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) ...
- 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 ...
- 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 ...
- 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) ...
- 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 ...
- 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
- 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 ...
- 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 # ...
- IO 优化
转自 BlackJack_ #define fastcall __attribute__((optimize("-O3"))) #define IL __inline__ __at ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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/
- 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 ...
- 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 ...