general electric blender

system to obtain the feedback information of the motherland, so as to realize the self-discipline and the allocation of grasping force. "Dexterous Hand Adaptive Operation Flow" is: pre-set, contact, hold, squeeze and relax 5 states, transitions between states rely on the contraction and stretching of a single muscle to achieve C. The time domain characteristics (over 0 pips, average absolute value, etc.) of EMG signals in Canada using different modes of operation, using artificial neural networ

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

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.#include #include#includeusing namespaceStd;__int64 gcd (__int64 x, __int64 y) {returnx%y==0? Y:GCD

whether these numbers can be similar into the stack out of the operation of the stack to get the number of questions given to the series;It's good to simulate with a stack.PS: Be sure to pay attention to the output format!!! After writing the code, Leng has been WA, started because the stack did not empty, and then found that the output format is not correct!!!The lesson of blood, changed long time!!!1#include 2#include 3#include 4#include 5#include 6#include 7#include 8#include 9#include Ten#i

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-

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 be purchased for rice, and you can assume that you are spending more than all of the rice, and you c

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

, (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, assuming we just find out the P-state of A≤b. Then they are (0,0). (For each). (3,5), (4,7), (6,10) ... Do they have any rules? Ignoring (0,0), it is very fast to find a for the P-State of I. A = i

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

[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);}}}}printf"%d\n", Dis[max]);}intMain (){intN while(~SCANF ("%d", n)){Min = INF;Max =-inf; CNT =0;memset (Vis,0

22 31 3Sample Output0.5000.4000.500The shortest path used before is the Min function initialization array map[][] is positive infinity, the problem of maximum security to initialize the array with the MAX function is negative infinity.1#include 2#include 3 #defineInf-0xfffffff4 using namespacestd;5 intN;6 Doublemap[1005][1005];7 voidF1 ()8 {9 intk,i,j;Ten for(k =1; K ) One { A for(i =1; I ) - { - for(j =1; J ) the { -Map[i][j]=max (Map[i][j

intersections: 3 Parallel lines and 1 free line intersection number + 1 free line can form the number of intersections, that is 3+0=33, when i=2,n-i=2, the number of intersections: 2*2+{0,1}={4,5}4, when i=1,n-i=3, the number of intersections: 1*3+{0,2,3}={3,5,6}each free line has an intersection with each parallel line, and the number of intersections of the J free and I parallel lines is j*i, so the intersection of the n lines equals the intersections of the free and parallel lines plus the i

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,

[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]+Price[u][j]; }Else if(dis[u]+map[u][j]==dis[j]pre[u]+price[u][j]Pre[j]) pre[j]=pre[u]+Price[u][j]; }

multiple sets of input data. The first behavior of each group of data is a positive integer n (0OutputEach set of input data corresponds to one row of output. Output an integer m, indicating that Gameboy may receive a maximum of M pies.Tip: The amount of input data in the subject is relatively large, it is recommended to read in scanf, with CIN may time out.Sample Input65 14 16 17 27 28 30Sample Output4A number of tower models, from the bottom up, to add a coordinate, to find d[0][6], the array

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; -

; - } thescanf"%d",m); * for(i =0; I ) $ {Panax Notoginsengscanf"%d%d",a,b); -d[a]++; thed[b]++;//record the degree of each point + F1 (A, b); A } theCnt=0; key=0; + for(i =1; i) - { $ if(D[i]%2)//To judge that each point is an even degree $ { -key=1; - } the if(i = = Fa[i])//Judging there's no independent point - {Wuyicnt++; the } - } Wu if(CNT = =1)//There

1#include 2#include 3 intMain ()4 {5 intn,m,i,j,t,x,y;6 while(~SCANF ("%d%d", x,y) (x| |y))7 {8 for(i=x,t=1; ii)9 {Tenx=i*i+i+ A; OneN= (int) sqrt (Double) x); A for(j=2; jj) - { - if(! (%j)) the Break; - } - if(jN) -t=0; + } - if(t) +printf"ok\n"); A Else atprintf"sorry\n"); - } - return 0; -}Hangzhou Electric