tegra k1

;} + intMid= (LL+RR) >>1; ABuild (k1, ll,mid); theBuild (k1|1, mid+1, RR); +sum[k]=sum[k1]+sum[k1|1],bo[k]=bo[k1]bo[k1|1]; - } $LL Query (intKintllintRR) $ { - if(L[K]GT;=LLAMP;AMP;R[K]returnSum[k]; - intMid= (L[k]+r[

name.ljust (20, ' * ') 52 # Content right-aligned rjust (width, padding) name.ljust print name.lower (58, ' '), print name.upper (), #变成小写lower55 #大小写交换swapcase59 name = "AleX" Print name.swapcase () #移除左边的空白lstrip62 print Name.lstrip () #分隔partition (front, middle, rear) 64 name = "Hello CGT", "Name.partition" ("ll"), #替换replace (old,new), the name = ' Alex ' of the print Name.replace (' A ', ' B ') #从右边 Start looking for RFIND70 print name.rfind (' e ') #以什么开始72 print name.startswith (' a '

1, the following variables (TU is a tuple), please implement the required functionsTu = ("Alex", [One, one, 11, {"K1": ' V1 ', "K2": ["Age", "name"], "K3": (, 22, 33)}, 44])0 10 1 20 1 2A. Describing the attributes of a tuplePrint ("Tuples are immutable lists, which can put data of any data type, can be queried, can be looped, can be sliced, but cannot be changed.")B. Can you tell me if the first element in the TU variable, "Alex", could be modified?P

A dictionary is another mutable container model and can store any type of object. Each key value of the dictionary (Key=>value) is split with a colon (:), each pair is separated by a comma (,), and the entire dictionary is enclosed in curly braces ({}) , as shown in the following format:1. Basic institutionsinfo = { "K1""v1"# Key-value pairs " K2 " " v2 " # Key-value pairs }2, the dictionary value can be any value#!/usr/bin/env pythoninfo = {

-load-truncated yes Lua-time-limit 5000 Slowlog-log-slower-than 10000 Slowlog-max-len 128 Latency-monitor-threshold 0 Notify-keyspace-events "" Hash-max-ziplist-entries 512 Hash-max-ziplist-value 64 List-max-ziplist-size -2 List-compress-depth 0 Set-max-intset-entries 512 Zset-max-ziplist-entries 128 Zset-max-ziplist-value 64 Hll-sparse-max-bytes 3000 Activerehashing yes Client-output-buffer-limit normal 0 0 0 Client-output-buffer-limit slave 256mb 64mb 60 # Client-output-buffer-limit pubsub 32m

, according to the requirements of each functionTu = (' Alex ', ' Eric ', ' Rain ')a. 计算元组长度并输出tu = (‘alex‘, ‘eric‘, ‘rain‘)print(len(tu))b. 获取元组的第 2 个元素，并输出tu = (‘alex‘, ‘eric‘, ‘rain‘)print(tu[1])c. 获取元组的第 1--‐2 个元素，并输出tu = (‘alex‘,‘eric‘,‘rain‘)print(tu[0:2])d. 请使用 for 输出元组的元素tu = (‘alex‘,‘eric‘,‘rain‘)for i in tu : print(i)e. 请使用 for、len、range 输出元组的索引tu = (‘alex‘,‘eric‘,‘rain‘)for i in range(len(tu)) : print(i)f. 请使用 enumrate 输出元祖元素和序号（序号从 10 开始）tu = (‘alex‘, ‘eric‘,

#include #include#include#include#includeusing namespaceStd;typedefLong LongLL;Const intN = -+2;Const intMoD =998244353;intt,n,m,lim,q,tot,w[n],dp[n][n][n][3],head[n];structedge{intV,next;} Edge[n*n/2];voidAddintUintv) {EDGE[TOT].V=v; Edge[tot].next=Head[u]; Head[u]=tot++;} InlinevoidUpintx,inty) {x+=y;if(X>=mod) x-=MoD;}BOOLJudgeintIintj) { returnABS (W[i]-w[j]) Lim;}intMain () {scanf ("%d",T); while(t--) {scanf ("%d%d%d%d",n,m,lim,q); for(intI=1; i"%d",W[i]); memset (Head,-1,sizeof(head)),

(intXintchain) { intk=0; SZ++; POS[X]=sz;//assigning the number of X nodes in the segment treebelong[x]=chain; for(intI=HEAD[X]; I I=e[i].next)if(deep[e[i].to]>deep[x]size[e[i].to]>Size[k]) k=e[i].to;//Select subtree Max son inherit heavy chain if(k==0)return; DFS2 (K,chain); for(intI=HEAD[X]; I I=e[i].next)if(deep[e[i].to]>deep[x]k!=e[i].to) DFS2 (e[i].to,e[i].to);//The rest of the sons new open heavy chain}intLcaintXintY//Seeking LCA{ if(deep[x]Deep[y]) swap (x, y); intt=deep[x]

Advanced Data StructuresOne, left bias tree oblique heapMerge, INSERT, deleteHit MarkstructNode {intFA,L,R,W,DEP} TREE[MX];intMerge (intK1,intK2)//The return value is the root node{ if(k1==0|| k2==0)returnk1+K2; if(k1.valk2.val) Swap (K1,K2); TREE[K1].R=Merge (TREE[K1].R

straight line and used as the other two sides of the rectangle. 5. traverse all adjacent two points of the convex hull and run the new two ~ 4. Use the rectangle with the smallest area as the result. Normally, the rectangle goes through five random points. The result is as follows: The Matlab code is as follows: Clear all; close all; clc; n = 30; P = rand (n, 2); ind = convhull (P (:, 1), P (:, 2 )); L = length (IND); hull = P (IND, :); % random dot Convex Hull area = inf; for I = 2: l p1 = hu

]) print (obj) ########## Result:counter ({22:2, 33:1, 11:1}) Counter ({22:3, 11:2, 33:1, 44:1}) Counter ({22:2, 33:1, 11:1, 44:0})Second, ordered dictionary (ordereddict)An ordered dictionary (ordereddict) is a subclass of a dictionary that records the order in which elements are added, based on a dictionary#导入collections模块import Collectionsdic = collections. Ordereddict () dic[' k1 '] = ' v1 ' dic[' k2 '] = ' v2 ' dic[' K3 '] = ' v3 ' Print (DIC) ##

1.hash type OperationImport Redispool = Redis. ConnectionPool (host= "192.168.48.131", port=6379, db=0) R = Redis. Redis (Connection_pool=pool)#hash类型操作: is a name corresponding to a dictionary#语法 Hset (name, key, value)A key-value pair is set #name the corresponding hash (does not exist, the key-value pair is created; otherwise, the key-value pair is modified)# syntax parameter explanation:Name:redis's hash nameKey:key1 corresponding to the key in the hashValue:value1 value in the corresponding

original value is replaced with the new value. Test code: [Java]Package test. junit;Import java. util. HashMap;Import junit. framework. TestCase;/*** In HashMap, determine whether two keys are the same: first judge the hashCode and determine whether equals* E. hash = hash (k = e. key) = key | key. equals (k ))* @ Author xuefeng**/Public class HashMapTest extends TestCase {/*** The hashCode of v1 and v4 is the same, but the equals does not match. Therefore, v1 and v4 form a linked list.*/Publi

T1:"Data Range"40% of the data meet a;another 30% of the data meet n,m;| s|,| T|100% of data meet n,m;| s|,| t|. The number of occurrences of the match in the Loop section, that is, the LCM (S,t), is first obtained.Then multiply the number of loops.is the number of equal character logarithm of the position of D=GCD (s,t) in the s,t.Can prove.Prove:Set S length to l1,t length of L2If the character of position A in S is the same as the character of position B in T.If it can be matched within the l