funny ladder

In the past few days, SINA/SOHU/163 has been connected to CSRF .. CSRF is not valued in China. Why is it funny? Because the form of the link is protected by CSRF, it is actually... You can modify the mailbox settings and use the GET method. Therefore, the attacker can directly send an email to the recipient, and the recipient will immediately win the bid as soon as he sees the email. In my demo, all user emails are forwarded to a specified email addr

cases to draw conclusions.Depending on the condition, the possible occurrence of the position I (i>0) is: Assuming that I-bit is known Number of previous I-bit occurrences(The other has not appeared before the first digit) 0 Because the first bit is 2, so there must be 2. 0,1 A[i] 2 A[I+1] = A[i] B[i] 0,2 B[I+1] = B[i]+b[i]+a[i] C[i]

DescriptionThere is n+1 rows and m+1 columns fence with n*m grids on the grassland. Each of the grids has a sheep. In order to let the sheep together, we need to dismantle the fence. Every time you can remove a row or a column of fences. What's the least number of times to reach the goal?InputThere is multiple test cases.The first line of all case contains, integers N and M. (1≤n,m≤1000)Outputfor each case, the output of the answer in a line.Sample Input1 22 2Sample Output12The main topic: a n*m

This time I did not do it, see a lot of people including seniors code to understand, I feel the best method or the following one, the mark with whom the beginning is not good, I feel a bit difficult to understand, if you do not understand the example of the group on the paper will be more clear#include #include#includeusing namespacestd;#defineLL Long LongLL a[500001];BOOLflag[500001];intMain () {intT,n,i,total,id=1; LL sum; scanf ("%d",t); while(t--) {scanf ("%d",N); memset (Flag,0,sizeof(f

money. If left-leaning (i.e. Abe Heavy, CDI light). The first time it was called, Abcd>efgh. So the counterfeit money in AB, and the counterfeit currency heavy; A and I weigh once, xxiv B is counterfeit money, if a heavy, then a is counterfeit money. If the right (ie, Abe Light, CDI Heavy). The first time it was called, Abcd>efgh. So the counterfeit money is in CD, and the counterfeit currency is heavy; C and I weigh once again, XXIV D i

Problem Descriptionalex have invented a new game for fun. There isNIntegers at a board and he performs the following moves repeatedly:1. He chooses three numbersa,bandCWritten at the board and erases them.2. He chooses the numbers from the triplea,bandCand calculates their greatest common divisor, getting the numberD(DMaybegcd(a,b) ,gcd(a,c) Orgcd(b,c) ).3. He writes the numberD to the board.It can be seen this after performing the move n−2 times, there'll be is only a numbers Wit

//_ooooo_//o8888888o//88 ". "The// (| -_- |)//o\ =/o//____/'---' \____// . ' \\| |// `.// / \\||| : |||// // / _||||| -:- ||||| - // | | \\\ - /// | |// | \_| ' \---/' | |// \ .-\__ `-` ___/-. ///___ '.. '/--.--\ ' ... __// ."" ' // | | : `- \`.; `\ _ /`;. `/ - ` : | |// \ \ `-. \_ __\ /__ _/ .-` / ///

It's not funny to get into a crowded car today. Generally, there are many people at peak hours in the morning, and many people prefer to use backdoors. When I got on the bus, I hurried up to four mm. When I got on the bus, everyone paid for me (I usually stood in the middle of the throat and liked the feeling of collecting money, hehe, kidding me ), let me upload it. Although the quality of Guangzhou residents is generally high, there are still fre

Topic Connection: http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1333Test instructions: give you an n vertex, M edge (each edge has three parameters, opening time, closing time, and the time of passing this edge) of the direction graph;You need to find the shortest path from S to T; Dijkstra algorithm can be solved;Pit point: I use the queue optimization +vector store each edge, after each call Dijkstra, must initialize the adjacency table, in this place for a long time, this is the second, af

More specifically, this article should be the programmer in the eyes of IE browser, I believe if you are a programmer, should not choose to use IE browser, if you are still using Internet Explorer, should not be a programmer (haha ... A little joke). Okay, let's just say that, and then we'll take a few pictures to see what the browser looks like in a programmer's eye:If the browser is a gunThe idea is a little different how to differentiate between HTML and HTML5 growing annoyance of the mainstr

I have been in the past 24 hours, has been the problem of the toss ...Topic:A Math GameTime limit:2000/1000ms (java/others) Memory limit:256000/128000kb (java/others)Problem DescriptionRecently, Losanto find an interesting Math game. The rule is Simple:tell your a number H, and you can choose some numbers from a set {a[1],a[2],......, a[n]}. If the sum of the number you choose are H, then you win. Losanto just want to know whether he can win the game.InputThere is several cases.In each case, the

} * $ voidDijkstraints) {Panax Notoginseng for(intI=1; iINF; -d[s]=0; thememset (Used,0,sizeof(used)); + A for(intk=1; k){ the intu,m=INF; + for(intI=1; iif(!used[i]d[i]i]; -used[u]=1; $ for(inti=first[u];i!=-1; i=Next[i]) { $ intv=e[i].v,a=e[i].a,b=e[i].b,t=e[i].t; - - if(aContinue;//The open time is less than the time passed the - intnow=d[u]% (A +b);Wuyi if(now+t//now able to p

Title: http://poj.org/problem?id=2484Test instructions: N a stone around a circle, two people take turns, each time can take a stone or adjacent to the two stones, ask the first win or the next winAnalysis:A typical symmetric gameIf n>=3, then no matter how the choice, the hand can choose a way to make the remaining two is the same length of the chain, then regardless of how the operation of the next, as long as in the later game to take the same symmetrical operation, will be able to let the op

Question link: http://acm.csu.edu.cn/OnlineJudge/problem.php? Id = 1333 Solution Report: There are n vertices and M unidirectional edges in a graph. Note that they are unidirectional edges, and then turn off B s for a second for each drive, the shortest time from S to T. A simple Shortest Path changes slightly. It took a long time because the side was not seen as one-way side and speechless. You can use queue optimization. 1 # include View code CSU 1333 fun