lenovo 1421

June 5, Lenovo held in Beijing Lenovo Z5 New National flagship conference, officially released Lenovo Z5, Lenovo K5 Note, Lenovo A5, Lenovo Watchⅹ, particle mineral cloud and other new products.This is

1421 Max MoD valuetitle Source: Codeforcesbase time limit: 1 seconds space limit: 131072 KB score: 80 Difficulty: 5-level algorithm problemThere is an an an array of n integers. Now you want to find two numbers (can be the same one)ai,aJ Makesai mod aJ Max and ai ≥ aJ. InputA single set of test data. The first line contains an integer n, which represents the size of the array A. (1≤n≤2*10^5) The second line has n spaces separated by an integ

http://acm.hdu.edu.cn/showproblem.php?pid=1421 Chinese questions Idea: First to find out how to combine all the items into a pair of, the problem requires the smallest fatigue, that is, a pair of two items in the weight of the smallest difference. Then you need to use the sorting, the items from small to large, from traversing the items until the second to the last, each item and the next item is combined into a pair, it will be found that if we first

Cisco IP 8800 device Web application Denial of Service Vulnerability (CVE-2016-1421)Cisco IP 8800 device Web application Denial of Service Vulnerability (CVE-2016-1421) Release date:Updated on:Affected Systems: Cisco IP Phones 8800 Description: CVE (CAN) ID: CVE-2016-1421Cisco IP 8800 Series Phones is a digital telephone system product.In Cisco IP 8800 Series Phones 11.0.1 and earlier versions, the We

Label: style blog HTTP Io OS AR for SP Div Question address: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1421 N items are given. You need to select K pairs of items to minimize the sum of the squares of each pair of items. Thinking; Dynamic Planning and solving, first sort items by weight in ascending order. According to greedy ideas, each pair of items must be two adjacent items, so that the square of the weight difference can be minimized. Then,

Linux Kernel Local Denial of Service Vulnerability (CVE-2015-1421) Release date:Updated on: Affected Systems:Linux kernelDescription:Bugtraq id: 72356CVE (CAN) ID: CVE-2015-1421 Linux Kernel is the Kernel of the Linux operating system. Linux kernel has the net: sctp: slab vulnerability caused by re-exploitation after INIT conflicts are released. Local attackers can exploit this vulnerability to cause sy

Meaning... First, sort the order from small to large, and obtain it after analysis. If you want to go to the I, then the I-1 item also need to take (because it is out of order and I difference is the smallest I-1 or I + 1, But I + 1 and I can also be seen as I and I-1, so if you want to go to the I, then the I-1 item also needs to take ). Analysis: Set DP [I] [J] to indicate that there are I items, and use j pairs. The I-th item has two conditions for DP [I] [J: 1: If you do not want to take the

Tags: des style blog HTTP Io color OS ar Java Move dormitory Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others) Total submission (s): 17885 accepted submission (s): 6065 Problem description it is very tiring to move the bedroom, xhd deep experience. time to record July 9, 2006, xhd was forced to move from building 27 to building 3 that day, because building 10 had to be closed. looking at the N items in the dormitory, xhd started to get in a daze. Because N is an

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1421 1 #include View code

The first thing you can see is a conclusion: it is definitely the number that is adjacent to each selection after the sort is finished.DP[I][J] indicates that the minimum cost of the J Group is selected for the first I number.#include #include#include#include#includeusing namespacestd;Const Long Longinf=999999999999999;Const intmaxn=2200;intn,k;Long LongA[MAXN],DP[MAXN][MAXN];intMain () { while(~SCANF ("%d%d",n,k)) { for(intI=1; i"%lld",A[i]); Sort (a+1, A +1+N); for(intI=0; i) for(

Obviously, after the items are sorted by weight, if you want to move a pair of items, these two items must be adjacent.Then the order of the first item to consider the article I put or not put, that is to get the state transfer equation:F[i][j] = min (f (i-1, J), F (i-2, j-1) + d[i]);Among them, D[i] indicates the fatigue caused by the pair of articles I and article i-1.1#include 2#include 3#include 4 using namespacestd;5 6 Const intN =2001;7 Const intM =1001;8 intF[n][m];9 intC[n];Ten intN, K;

line of data that represents his minimum fatigue.Sample Input2 11 3Sample Output41#include 2#include 3 using namespacestd;4 5 Const intinf=0x3f3f3f3f;6 intdp[2005][1005];7 intv[2005];8 9 intMain ()Ten { One intn,k,i,j; Adp[0][0]=0; - while(SCANF ("%d%d", n,k)! =EOF) - { the for(i=1; i) -scanf"%d",v[i]); - for(i=0; i) - for(j=1; j) +dp[i][j]=inf; -Sort (v,v+n+1); + for(i=2; i) A { at for(j=1; j*2) -Dp[i][j]=min (dp[i-2][j-1]+

Move BedroomTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 18571 Accepted Submission (s): 6290Problem description Move bedroom is very tired, XHD deep have experience. The date of the July 9, 2006, the day Xhd forced to move from building 27th to building 3rd, because 10th to seal the building. Looking at the bedroom n items, XHD began to Daze, Because n is an integer less than 2000, it is too much, so XHD decided to move 2*k pieces of the p

#1421: Four fork tree time limit: 20000ms single point limit: 2000ms memory limit: 256MB description Little ho: Next weekend we're going to play in the next city. Small hi: Anyway, very close to the back, good ah. Little ho: Well, let me start by planning the route. Little Ho opened the map app in the phone and moved the coordinates to the city next door. A variety of shops are displayed on the street map. Little hi: Little ho, do you know how the map

Wust oj 1421 we love girl (greedy or DP)DescriptionWhen taking part in ACM Programming Contest, middle school hope girls for registration like ccnu, cug and so on. so this year our wust send more girls as possible to have a partition tion. the ratio of male to female of wust is seven to one, so we may not have so far girls. of course the rest exceptionist is boys.Now there are n schools take part in Wust ACM Programming Contest, x girls and y boys is

#include 　　HDU 1421 moving Dorm (DP)

named Dp,dp[n][k] is to take K pairs from the n number, and then use a one-dimensional array to deposit input values (to sort).When N==2*k, dp[n][k] = dp[n-2][k-1] + (S[n]-s[n-1]) ^2 because all the numbers are taken, there is no need to compare anything.When n > 2*k, dp[n][k] = min (dp[n-2][k-1] + (S[n]-s[n-1]) ^2,dp[n-1][k]) either take nth and n-1 to form a pair, or discard the nth re-count.1#include 2#include 3#include string.h>4 #defineMAX 20055 #defineINF 21474836406 7 intDp[max][max/2];8

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1421 Analysis: to minimize the sum of squares of the difference, the two pieces obtained at a time must be the two that are placed together after sorting. Therefore, the array is first sorted. if I = 2 * j, select all. If I add a number (I + 1), select either of the two conditions. that is: DP [I] [J] = min (DP [I-1] [J], DP [I-2] [J-1] + C ); C = (F [I]-f [I-1]) ^ 2; That is to say, i

#include #include #include #include #include #include #include #include #include #include using namespace STD;#define INF 0X2FFFFFFF#define LL Long Long#define MAX (A, B) ((a) > (a))? (a):(B)#define MIN (a) inta[2005];intdp[2005][10005];intMain () {intN,m; while(Cin>> n >> m) { for(inti =1; I scanf("%d", a[i]); } sort (A +1, a+n+1); for(inti =0; I 0][i] = INF; dp[1][i] = INF; } for(inti =0; I 0] =0; } for(inti =2; I for(intj =1; J if(I 2*J) {Dp[i][j] = INF; }Else{Dp[i][j] = min (dp[i-2][j-1]+ (

, poor xhd wants to know what the best status (that is, the lowest fatigue) will be after these 2 * k items are moved. Please tell him. Each input group has two rows. The first row has two numbers, N and K (2 Output corresponds to each group of input data, and only one output data indicates its minimum fatigue, each row. Sample input2 11 3 Sample output4 Authorxhd Source ACM summer training team exercise session (2) transfer equation: DP [I] [J] = min {DP [I-1] [J], DP [I-2] [J-1] + (AA [I]-aa [