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[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
)Sample Inputsosoonrivergoesthemgotmoonbeginbig0 Sample Outputyes. Harry can read this mantra: "Big-got-them."#include #include#include#includestring.h>using namespacestd;BOOLvis[ +];Charstr[ +][ +]; BOOLBFsintN) {Queueint>que; intA, Len; for(inti =0; I //first, the first letter is a b queue. { if(str[i][0] =='b') {Que.push (i); Vis[i]=true;//indicates that the word has been marked. } } while(!Que.empty ()) {a=Que.front (); Que.pop (); Len= strlen (Str[a])-1;//The
1#include 2 intMain ()3 {4 inta[ -],n,i,t,k;5 while(~SCANF ("%d", n) N)6 {7 for(i=k=0; ii)8 {9scanf"%d",a[i]);Ten if(A[i]//looking for the minimum value position Onek=i; A } - if(k)//not the first one, then the first exchange. - { thet=a[0]; -a[0]=A[k]; -a[k]=T; - } + for(i=0; ii) - if(i)//Control output Format +printf"%d", A[i]); A Else atprintf"%d", A[i]); -printf"\ n"); - } - ret
); This is the prototype, which decomposes the string into a set of strings. The string that s wants to break down. D is a delimiter character because all D is substituted for the first execution of this function, so you need to change s to NULL when you call it again (the C language takes it as the last character of the string to represent the end of the string. ") ").Function Three:Atoi () (representing alphanumeric to integer) is a function that converts a string into an integer number. belon
The 1004 question is actually very simple, is wants you to count the most balloons the number, but in the output time but needs to be careful carefully, conforms to the question request.In the output is the first note is the output order: If there are multiple numbers of the same color, first input first output, such as,RedGreenRedGreenThis situation requires an outputRedGreenPost the AC code:#include Hangzhou Electric OJ 1004 realization and some t
problem of precision.See the code for details:1#include 2 intMain ()3 {4 inth,m,s,a,da,tots= A*3600* One, kase=1;5 while(SCANF ("%d:%d:%d", h,m,s) = =3)6 {7 intTar = One* (h*3600+m* -+s);8scanf"%d",a);9Da = the-2*A;Ten intCNT = A * -, F =1; One //The position at the beginning is a° (obviously the first minute is the second hand in front) . A while(cnttar) - { - if(f) the { -f=0; -cnt+=da* -;//The first Chase is the an
descriptiongiving N integers, V1, V2,,,, Vn, you should find the biggest value of F.Inputeach test case contains a single integer N (11#include 2#include 3 using namespacestd;4 inta[100000];5 intMain ()6 {7 __int64 n,j;8 while(SCANF ("%i64d", n) N)9 {Ten for(__int64 i=0; i) One { Ascanf"%d",a[i]); - } -Sort (a,a+n); thej=a[n-1]*( .-n); - for(__int64 i=1; i ) -j=j+a[n-1-i]; -printf"%i64d\n", j); + } -}Hangzhou El
Descriptiona range is given, the begin and the end are both integers.You should sum the cube of all the integers in the range. Inputthe first line of the input was T (1 Each case of input is a pair of integer A, a (0 1#include 2__int64 a[1000000]={0,1};3 intMain ()4 {5 for(__int64 i=1; i1000000; i++)6 {7a[i]=a[i-1]+i*i*i;8 }9__int64 t,k=0;;Tenscanf"%i64d",t); One while(t--) A { - __int64 m,n; -scanf"%i64d%i64d",m,n); theprintf"Case #%i64d:%i64d\n", ++k,a[n]-a[m-1]); - }
) is multiplied, it overflows. The title asks "may assume the result is in the range of 32-bit signed integer", which requires that the sum is a 32-bit signed integer. The sum of the test data given by OJ (n (n+1)/2) must be within the range of 32-bit integers, but (n (n+1)) is not necessarily. The reason you can infer WA should be here. You can subtly change the formula:Putsum=n* (n+1)/2;Switchif (n%2==0)sum=n/2* (n+1);ElseSum= (n+1)/2*n;The entire code is as follows:1#include 2 intMain ()3 {4
=getchar ())if(ch=='-') f=-1; for(; isdigit (ch); Ch=getchar ()) x=x*Ten+ch-'0'; returnx*F;} InlinevoidWriteintx) { if(x0) Putchar ('-'), x=-x; if(x>9) Write (x/Ten); Putchar (x%Ten+'0'); return ;}intn,m;inthead[100006];intnxt[200006],to[200006];intTotal=0;voidAddintXinty) { Total++; To[total]=y; Nxt[total]=Head[x]; HEAD[X]=Total ; return ;}intdfn[100006],low[100006];inttot=0;intnum[100006];voidTarjan (intXintFA) {Dfn[x]=low[x]=++tot; for(intE=head[x];e;e=Nxt[e]) { if(!Dfn[to[e]])
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