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++; Do{j = S.top ();S.pop ();INSTACK[J] =0;} while(J! = x);}}voidSolve (){Vis_num = CNT_SCC =0; for(inti =1;I if(Tim[i] = =-1)Tarjan (i);if(CNT_SCC = =1)Puts"Yes"); ElsePuts"No");}voidGetmap (){intA, B;memset (Instack,0,sizeof(Instack));memset (Low,0,sizeof(low));Memset (Tim,-1,sizeof(Tim));CNT =0;Memset (Head,-1,sizeof(head)); while(m--){scanf"%d%d", a, b);Add (A, b);}}intMain (){ while(~SCANF ("%d%d", n, m), n+m){Getmap ();Solve ();}return0;}Mygod1 Great God Analysis (Scc_tarjan): 2 http://blo

Outputyes.HintHintharry can read this mantra: "Big-got-them". This problem gives me a lot more of the flexibility of global variables use global variables for values that have an effect on the main function in the calling function and then the understanding of backtracking requires that the searched node also needs to be searched and that the topic does not need to be written to the DFS call function (the alignment is fixed) The first thing to be clear is the condition of the recursive terminat

multilayer perceptron as classifier to identify different muscle contraction patterns in 4. But its disadvantage is that the grasping mode and motion control of prosthetic hand still need the visual feedback of human eyes. Hybrid control Method: If the methods of B and C are combined, using C to determine the capture mode in B, and then adopt the adaptive operation process, the crawl performance will improve. 4. Algorithm based on pattern recognition Feature classification--Smart Hand Controlle

Problem Descriptiongiven A string containing only ' a '-' Z ', we could encode it using the following method:1. Each sub-string containing k same characters should being encodedto "KX" where "X" is the only character in this Sub-strin G.2. If the length of the sub-string is 1, ' 1 ' should beignored.Inputthe first line contains an integer N (1 Outputfor each test case, output the encoded string in Aline.Sample INPUT2ABCABBCCCSample outputabca2b3c #include #include #include using namespace Std;in

inputs, outputs a row, if there is a maximum not available score, output this score, otherwise the output infSample INPUT3 72 20 0Sample Output11infSourcethe 6th UESTC Programming ContestRecommendlcy | We have carefully selected several similar problems for you:1573 1060 1018 2535 3244 n days without knocking over the title, my fault. Brain Tease TM fast out of Xiang. First of all, look for a two times the feeling. Find the law, see the puzzle. Summarize it. int doesn't know why he can't live.#

; - while(bt) - { in if(S[a] = =E[b]) - { toa++;b++; +code[c++]=1; -code[c++]=0; the } * Else if(!sk.empty () e[b] = =sk.top ()) $ {Panax NotoginsengSk.pop (); b++; -code[c++]=0; the } + Else if(A t) A { theSk.push (s[a++]); +code[c++]=1; - } $ Else $ { -Flag =0; - Break; the } - }Wuyi if(flag) the

Max Sumproblem Descriptiongiven A sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a Sub-sequenc E. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.Inputthe first line of the input contains an integer T (1Outputfor Each test case, you should output of the lines. The first line was "Case #:", # means the number of the the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-

, they give us knowledge, teach us to be a manThank the friends, they let us feel the warmth of the world;Thanks to the opponents, they make us keep making progress and efforts.Likewise, we would like to thank the pain and hardship brought to our wealth ~ Input data first contains a positive integer c, which indicates that there is a C set of test cases, the first line of each set of test cases is two integers n and m (1Output for each set of test data, export the maximum weight that can b

Problem Description this summer holiday training team The first composition girls team, including a team called RPG, but as one of the training team members of the wild camel unexpectedly do not know RPG three person specifically who. RPG gave him the chance to let him guess, the first guess: R is the princess, p is the grass, G is the Moon hare; the second guess: R is grass, p is the Moon hare, G is the princess; Third guess: R is grass, p is princess, G is moon Hare; The poor wild camel finall

win the state, you face this situation will certainly win, just to follow the rules to take once you can. Then see y = the lattice point that is not crossed above X, which is P-state.K > 2 o'clock, (1,k) is not P-state. For example, if you face the 1,3 situation, you are likely to win. Similarly, (k,2), (1 + k, 2 + K) are also not P-states, and these points are zoned out and their symmetry points. Then find the remaining points above y = x, and you will find that (3,5) is a P-state, so go on, a

This kind of problem with decimals always gives me extra headaches./*id:modengd1prog:fence9lang:c++*/#include 　　Usaco Electric Fence

=0x3f3f3f3f;intdis[50010], vis[50010], Min, Max;structEdge{int from, to, Val, next;} edge[50005*3];inthead[50005], CNT;voidADD (intAintBintC{Edge E = {A, B, C, head[a]};EDGE[CNT] = E;Head[a] = cnt++;}voidSPFA (){ for(inti = Min;I Dis[i] =-inf;Dis[min] =0; The longest road;queueint> Q;Q.push (Min); while(! Q.empty ()){intU = Q.front (); Q.pop ();Vis[u] =0; for(inti = Head[u]; I! =-1; i = edge[i].next){intv = edge[i].to;if(Dis[v] {DIS[V] = Dis[u] + edge[i].val;if(!vis[v]){VIS[V] =1;Q.push (v);}}}

infinity + } - } the for(i =1; I ) * { $ for(j =1; J )Panax Notoginseng { -scanf"%LF",a); the if(I j) + { AMap[i][j]=map[j][i]=a;//record the security of every two points, in order to simply record the situation of I the } + } - } $ F1 (); $scanf"%d",q); - while(q--) - { thescanf"%d%d",b,c); - if(Map[b][c] >0)//if the two points are diffe

{ - for(k =0; K Free*( Free+1)/2; k++) - { the if(d[ Free][k] = =1) - { -d[i][ Free* (I- Free) +k]=1;//free* (i-free) +k indicates the number of intersections of the free straight line of the I line (k is the number of intersections of the free line) - } + } - } + } A while(SCANF ("%d", n)! =EOF) at { - for(i =0; I 1)/2; i++) - { - if(i = =0) -p

Xiao Ming A+bTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 37194 Accepted Submission (s): 17644Problem Description Ming is 3 years old, now he has been able to recognize 100 of non-negative integers, and be able to do a non-negative integer within 100 of the addition calculation.for integers greater than or equal to 100, xiaoming only retains the last two digits of the number for calculation, and if the result is greater than or equal to 100,

],PRICE[MX][MX],DIS[MX],PRE[MX];voidinit () { for(intI=1; i) for(intj=1; j) {Map[i][j]=INF; PRICE[I][J]=INF; }}voidDijkstraintv0) { BOOLVIS[MX]; for(intI=1; i) {Dis[i]=Map[v0][i]; Pre[i]=Price[v0][i]; Vis[i]=false; } Dis[v0]=0; Vis[v0]=true; for(intI=1; i){ inttmp=inf,p=Inf,u; for(intj=1; j) if(!vis[j]dis[j]tmp) {u=J; TMP=Dis[j]; } Vis[u]=true; for(intj=1; j) if(!Vis[j]) { if(dis[u]+map[u][j]Dis[j]) {Dis[j]=dis[u]+Map[u][j]; PRE[J]=pre[u

returnmax0; - } - intMain () - { + intN; - while(SCANF ("%d", n) N) + { A inta,b,max0=0; atmemset (D,0,sizeof(d)); - for(inti =0; I ) - { -scanf"%d%d",a,b); -d[b][a+1]++;//a pie that represents the second B's in the a+1 position. - if(b > Max0)//record the maximum value of the time in { -max0=b; to } + - } the for(inti = max0-1; I >=0; i--) * { $ for(intj =1; J O

Title: Portal.#include #include#include#includeusing namespacestd;intgcdLong LongALong Longb) { if(!B)returnA; returnGCD (b,a%b);}inta[100005];intMain () {intT,n; scanf ("%d",T); while(t--) { Long Longsum=0; scanf ("%d",N); for(intI=0; i) {scanf ("%d",A[i]); if(a[i]0) A[i] =-A[i]; Sum+=A[i]; } Long LongAP = Sum,aq =N; Long Longg =gcd (n,sum); AP= ap/G; AQ= aq/G; //printf ("%i64d/%i64d\n", Ap,aq); Long Longz=0; Long Longm=0; for(intI=0; i) {Z+=a[i]*A[i]; M+=2*A[i]; }

Title: Portal.Test instructions: A large number of N, up to 5 times the root of the root, ask several square root can get 1, if 5 times can not get 1 on the output tat.Problem solving:x1=1,x2= (x1+1) * (x1+1)-1, etc. X5 is not more than a long long , the output can be judged.#include #include#include#includeusing namespaceStd;typedefLong Longll;Charc[1005];intMain () { while(cin>>c) {intlen=strlen (c); if(len>= the) {printf ("tat\n"); } Else{ll n=c[0]-'0'; LL Maxx=4294967295; for(intI=1;

1#include 2 Chars[Wuyi];3 intLenth (Chars[])4 {5 intI=0;6 while(S[i])7++i;8 returni;9 }Ten intMain () One { A intI,n; - while(~SCANF ("%d",N)) - { the GetChar (); - while(n--) - { - gets (s); + intlen=Lenth (s); - for(i=0; ii) + { A if(i) at { - if(! (s[i]=='_'|| (s[i]>='a's[i]'Z')|| (s[i]>='A's[i]'Z')|| (s[i]>='0's[i]'9'))) - Break; -