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Dating with girls (2)Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 2416 Accepted Submission (s): 690Problem descriptionif You has solved the problem Dating with girls (1). I Think you can solve this problem too. This problem was also about dating with girls. Now you are in a maze and the girl you want to dat

ObjectiveIf you think of an online dating site looking for a date, you're likely to classify all users of the dating site as three categories:1. Do not like the2. A bit of glamour3. Very attractiveHow do you decide which category A user belongs to? Presumably you will analyze the user's information to get a conclusion, such as the user "frequent flyer miles per year", "playing online games consumes more tim

Dating with Girls (2) Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 2659 Accepted Submission (s): 744 Problem Description If You had solved the problem Dating with girls (1). I Think you can solve this problem too. This problem was also about dating with girls. Now you are in a maze and the girl you want t

Dating with Girls (2) Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 1476 Accepted Submission (s): 419 Problem Description If You had solved the problem Dating with girls (1). I Think you can solve this problem too. This problem was also about dating with girls. Now you are in a maze and the girl you want to

Now the online dating industry has become one of the largest economic sources of society. Almost 0 of the cost will make money, of course, this "0" is a little more than you think-you need to build a perfect site for future work.What do you think is the perfect website? Now I will explain to you the most valuable elements of such a site, but also for you to accurately understand each stage of the development process, so please do not miss this article

[] The author's views may not be all correct. But this articleArticleThe reality is worth thinking about.In fact, this blind date is also a form of dating, but there are too many people involved. While ignoring the real protagonist.In fact, similar "blind dates" often exist in the city ". It is often inferior to rural areas to evaluate the quality of human beings.What is the difference between money, status, education, and cars in the city and the hou

male Dating Is nothing more than a few points, but what are the points? small make up today for everyone to tidy up, do not really want to date, even this most basic points have forgotten. After reading this article, just remember these basic skills. 650) this.width=650; "Src=" Http://s5.51cto.com/wyfs02/M01/84/29/wKiom1eHMl6x2SnGAAFlAb7qHuY030.jpg-wh_500x0-wm_3 -wmp_4-s_2219734871.jpg "title=" simple and practical

Introduction and derivation of dating-number theory Extended Euclidean Algorithm for POJ-1061 frogs DescriptionThe two frogs met each other on the Internet. They had a good chat, so they thought it was necessary to meet each other. They are happy to find that they live on the same latitude line, so they agreed to jump westward until they met each other. However, before they set out, they forgot a very important thing. They did not know the characteris

is:%f"% (Errorcount/float (numtestvecs)) #输出错误率 - + if __name__=='__main__': ADatingclasstest () 1 " "2 Enter a person's information and give a forecast of their preferred level3 " "4 defClassifyperson ():5Resultlist = [' not at all','In small doses','In large doses']6Percenttats = Float (raw_input ("percentage of time spend playing video games?"))7Ffmiles = Float (Raw_input("frequent flier miles earned per year?"))8Icecream = Float (raw_input ("liters of ice cream consumed per year?"))9

$ (' # Cfmonitorid '). DataGrid (' LoadData ', data.rows);//Set multiple times clearinterval (Monitor_cf_interval); $ (' #returnMsg_monitorcf '). HTML (' Task run complete! ')///here to the background request to parse CF results, and storage, directly call CFRESULT2DB () function can cfresult2db ();} else{//set prompt and change page data, multi-line Display job task information//------set DataGrid Data $ (' #cfMonitorId '). DataGrid (' LoadData ', data.rows);}}); Backstage: After obtaining the

starting point of setting Frog A is x, and Frog B's starting point coordinates are Y. Frog A can jump M m at a time, Frog B can jump n meters at a time, and two frogs will spend the same time jumping once. Latitude line total length l m. Now you have to find out how many times they have jumped before they meet.InputThe input includes only one line of 5 integer x,y,m,n,l, where X≠y OutputOutput the number of hops required to meet, and output a line of "impossible" if it is never possible to meet

template. "TAT" number theory abuse me like a vegetable worm, even vegetable chicken can not forget ... AC Template ... /*/#include "map" #include "Cmath" #include "string" #include "Cstdio" #include "vector" #include "CString" #include "iostream "#include" algorithm "using namespace std;typedef long LL; ll EXGCD (ll a,ll b,ll x,ll y) {//Direct reference to save X and y in the entire process (N-M) x+ly=x-y; if (!b) {X=1;y=0;return A;} LL D=EXGCD (b,a%b,x,y); LL t=x;x=y;y= (t-a/b*y); return D;}

Title Link: http://poj.org/problem?id=1061is to find contentment (X+MT)-(Y+NT) = Lk's T and KThe upper formula can be reduced to (N-M) T + Lk = x-y, satisfying the form of ax+by=c so that we can seek t and K with the extension Euclid;Since the upper formula has a solution and only if C is gcd (A, B) = 0;#include #include#includestring.h>#includestring>#include#include#include#include#include#includeusing namespacestd;#defineMet (A, b) memset (A, B, sizeof (a))#defineN 10053#defineINF 0x3f3f3f3fC

2 have done, put 1 to fill.Test instructions gives N and K, and the number of N. is the n number to find the two number x, Y. Make x+y=k.Calculate all the possibilities. When X0+y0=k. X1+y1=k. Must have unequal. X0!=x1 or Y0!=y1.For example4 42 2 2 2The correct output should be 1.I was sort of, went heavy, and then two points. It seems that many people are using the pointers.Forget the STL's two-point return subscript function, the mood is not good, too lazy to check, hand hit two points.#includ

a time, Frog B can jump n meters at a time, and two frogs will spend the same time jumping once. Latitude line total length l m. Now you have to find out how many times they have jumped before they meet.InputThe input includes only one line of 5 integer x,y,m,n,l, where X≠y OutputOutput the number of hops required to meet, and output a line of "impossible" if it is never possible to meetSample Input1 2 3) 4 5Sample Output4#include #include#include#includeusing namespaceStd;typedefLong Longll;vo

[flag][v].ma = Tree[flag][v].ma_r = (tree[flag][v].r-tree[flag][v].l + 1) *tag;return;}Push_down (v);int mid = (tree[flag][v].l + tree[flag][v].r) >> 1;if (a if (b > Mid) update (A, B, right, flag, tag);Push_up (v);}int main (){Freopen ("In.txt", "R", stdin);int T;scanf ("%d", t);int cas = 0;while (t--){int N, Q;scanf ("%d%d", n, q);Build (1, N, 1);Char ch[10];int L, R, QT;printf ("Case%d:\n", ++cas);while (q--){scanf ("%s", ch);if (ch[0] = = ' D '){scanf ("%d", AMP;QT);if (Tree[1][1].ma {Puts (

; A=b; b=R; } returnA;}voidEXTEND_GCD (ll A, ll B, ll x, LL y) {if(b = =0) {x=1; Y=0; return ; } Else{EXTEND_GCD (b, a%b, x, y); LL tmp=x; X=y; Y= TMP-A/b *y; }}intMain () {LL x, y, D; while(~SCANF ("%i64d%i64d%i64d%i64d%i64d", x, y, m, n, L)) {LL a= N-M; LL b=L; LL C= X-Y; D=gcd (A, b); if(c% d! =0) {printf ("impossible\n"); Continue; } A/=D; b/=D; C/=D; EXTEND_GCD (A, B, X, y); LL T= c * x%b; if(T 0) T + =b; printf ("%i64d\n", T); } return 0;}View Code#include typedefLong Long

Topic Links:http://poj.org/problem?id=1061Main topic:Chinese topic, test instructions at a glance, is the data range is surprisingly large.Problem Solving Ideas:Assuming that two frogs have skipped T-times, you can list the indefinite equation: P*l + (n-m) *t = = x-Y. After the equation is listed, the solution of the indefinite equation is calculated using the extended Euclidean. In finding the smallest solution of the whole place card for a long time, long.To learn more about the use and proof

* * equation has an integer condition (y-x)%GCD (A, B). At this point the solution is extended gcd (y-x)/GCD (A, B). (Think about why?) Equivalent to both sides of the equation expand/Shrink k=(y-x)/GCD (A, B) times. but we need to ask for the smallest integer. Because the extended GCD is a set of integer solutions, this requires us to adjust the answer. The specific method isar = (ar% abs (b/g) + ABS (b/g))% abs (b/g); AR is the time to find out TYou can limit the answer to 0~abs (b/g) and wh

The processing of stones is based on the time of each location to obtain the remainder of K and judge the repetition. Others can be written at will. #include 　　 HDU 2579 dating with girls (2) BFS remainder judgment