题目链接: http://codeforces.com/problemset/problem/622/C 题目大意: 给定一个长度为n(n不超过200000)的序列,有m(m不超过200000)次询问,第i次询问一个区间[li,ri]内是否存在一个数不等于一个给定值x.如果存在,就输出这个数的位置,否则输出-1. 解题思路: 预处理一个数组p[n].p[i]表示从位置i向左一直到位置0,第一个不等于a[i]的数的位置.可以以o(n)的复杂度通过对递推实现.具体来说就是首先令p[0]=-1,然后从…

C. Not Equal on a Segment 题目连接: http://www.codeforces.com/contest/622/problem/C Description You are given array a with n integers and m queries. The i-th query is given with three integers li, ri, xi. For the i-th query find any position pi (li ≤ pi …

C. Not Equal on a Segment time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given array a with n integers and m queries. The i-th query is given with three integers li, ri, xi. For th…

C. Not Equal on a Segment time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given array a with n integers and m queries. The i-th query is given with three integers li, ri, xi. For th…

预处理p[i],p[i]表示:[p[i],i]这段闭区间上所有数字都是a[i] 询问的时候,如果xi==a[ri]并且p[ri]<=li,一定无解 剩下的情况都是有解的,如果xi!=a[ri],那么输出ri,否则输出p[ri]-1. 另外,看到有大牛博客说可以用线段树,大致是这样的: 线段树保存区间最大值与最小值, 如果询问的区间上最小值==最大值,那么无解: 剩下的情况都是有解:如果xi不等于最小值,那么输出最小值位置:如果xi不等于最大值,那么输出最大值位置. #include <stdi…

题目链接: http://codeforces.com/problemset/problem/622/C 题意: 给定序列,若干查询,每个查询给定区间和t,输出区间内任意一个不等于t的元素的位置. 分析: 最初没看样例直接钦定输出每个不等于t的元素位置,结果怎么想都是n2复杂度的,后来看了样例才发现是输出任意一个.. 对于一个区间,如果区间最大值和最小值相等,那么该区间元素值全部相同,那么我们维护区间的最大最小值,然后判断是否均等于t,若不等,输出最大值或最小值的位置即可,若相等, 则该区间所有…

622A - Infinite Sequence    20171123 暴力枚举\(n\)在哪个区间即可,时间复杂度为\(O(\sqrt{n})\) #include<stdlib.h> #include<stdio.h> #include<math.h> #include<cstring> #include<iostream> #include<algorithm> using namespace std; long long n…

安装过程: 1.configuration --prefix=PREFIX install all files under the directory PREFIX instead of usr/local/psql --with-pgport=NUMBER set NUMBER as the default port number for server and clients,the default is 5432. --with-segsize=SEGSIZE(控制表的大小) set the…

This document describes the installation of PostgreSQL using the source    code distribution. (If you are installing a pre-packaged distribution,    such as an RPM or Debian package, ignore this document and read the    packager's instructions instea…

Circles and Pi Introduction id: intro-1 For as long as human beings exist, we have looked to the sky and tried to explain life on Earth using the motion of stars, planets and the moon. The Greeks were the first to discover that all celestial objects…