at t 1040 manual

Topic links 1040: [ZJOI2008] Knight time limit:10 Sec Memory limit:162 MBsubmit:3054 solved:1162[Submit] [Status] [Discuss] DescriptionThe Knights of Z are a powerful organization that brings together elites from all over the world. They maxi and punish evil and Good, and are praised by all sectors of the community. A terrible thing has happened recently, and the evil Y countries have launched a war of aggression against Z-states. The

#1040: The rectangle determines the time limit:1000msSingle Point time limit:1000msMemory Limit:256MBDescribeThe 4 lines on the plane are given to determine whether the 4 segments are exactly enclosed in a rectangle with an area greater than 0.InputThe first line of input is an integer T (1Each set of data contains 4 rows, each containing 4 integers x1, y1, x2, y2 (0 OutputEach set of data outputs a row of Yes or no, indicating whether the input 4 seg

1040: [ZJOI2008] Knight time limit: ten Sec Memory Limit: 162 MB Submit: 1976 Solved: 772 [Submit] [Status] DescriptionThe Knights of Z are a powerful organization that brings together elites from all over the world. They maxi and punish evil and Good, and are praised by all sectors of the community. A terrible thing has happened recently, and the evil Y countries have launched a war of aggression against Z-states. The war s

1040: [zjoi2008] server guard time limit: 10 sec memory limit: 162 MB Submit: 1755 solved: 690 [Submit] [Status] Description The Knights of the Z state are a very influential organization, which brings together elites from all over the world. They have been praised by all walks of life for their part in the fight against wealth and poverty. A terrible thing happened recently. The evil country Y launched a war of aggression against the country Z. The

As easy as A+bTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 45268 Accepted Submission (s): 19340Problem Descriptionthese days, I am thinking on a question, how can I get a problem as easy as a+b? It is fairly difficulty to do such a thing. Of course, I got it after many waking nights.Give you some integers, your task was to sort these number ascending (ascending).Should know how easy the problem is now!Good luck!Inputinput contains multiple

Given a base ring forest, each vertex has the right value. to select a weight value and the largest vertex set, any two vertices in the vertex set cannot be directly connected. Largest vertex independent set ...... Considering n For each base ring tree, we find an edge on the ring and set the vertices at both ends of the edge to U and V respectively, f [I] is the maximum weight of the I-based subtree when I is taken, and g [I] is not. Therefore, we have the following practices: 1. Cut off this

1040. Longest shortric String (25), longestsymmetric Given a string, you are supposed to output the length of the longest distributed Ric sub-string. For example, given "Is PAT TAP distributed Ric? ", The longest specified Ric sub-string is" s PAT TAP s ", hence you must output 11. Input Specification: Each input file contains one test case which gives a non-empty string of length no more than 1000. Output Specification: For each test case, simply p

As easy as a+bTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) total submission (s): 53368 Accepted Submission (s): 22939problem Descriptionthese days, I am thinking on a question, how can I get a problem as easy as a+b? It is fairly difficulty to do such a thing. Of course, I got it after many waking nights. Give you some integers, your task was to sort these number ascending (ascending). Should know how easy the problem is now! Good luck!InputInput contains multip

Violence.#include #include#includeusing namespacestd;Chars[ -];intans,len,sum;voidCheckintAintb) { if(S[a]!=s[b])return; if(a==b) sum=1;Elsesum=2; intleft=a-1, right=b+1; while(! (left0|| right>=len| | s[left]!=S[right])) Left--, right++,sum=sum+2; Ans=Max (ans,sum);}intMain () {gets (s); Len=strlen (s); ans=0; for(intI=0; i1); printf ("%d\n", ans); return 0;}PAT (Advanced level) 1040. Longest symmetric String (25)

} + DFS (pur,x); - } the } * voiddpintXintFA) { $f[x][0]=0; f[x][1]=0;Panax Notoginseng for(intI=first[x];i;i=Next[i]) { - intPur=Go[i]; the if(b[i]==tmp| | PUR==FA)Continue; + DP (PUR,X); Af[x][0]+=std::max (f[pur][0],f[pur][1]); thef[x][1]+=f[pur][0]; + } -f[x][1]+=W[x]; $ if(flag==0x==root) f[x][1]=0; $ if(flag==1x==node) f[x][1]=0; - } - intMain () { thescanf"%d",n); - intId=0;Wuyi for(intI=1; i){ the intx; -scanf"%lld%d",w[i],x)

The string appapt contains two words "Pat", where the first Pat is 2nd (P), 4th (a), 6th (t), and the second Pat is 3rd (p), 4th bit (a), 6th (t).Now given a string, how many pat can it form?Input format:Input has only one row, contains a string, the length of not more than 105, only contains P, A, T three kinds of letters.Output format:Outputs the number of PAT contained in a given string in a row. Because the result may be larger, only the result of the remainder is output to 1000000007.Input

;voidDFS1 (intXintFA) {Vis1[x]=1; for(inti=head[x];i!=-1; i=Nex[i]) { if(to[i]!=FA) { if(!Vis1[to[i]) DFS1 (to[i],x); Else{St=x; En=To[i]; } } }}Long Longdp[maxn+5][2];intvis2[maxn+5];voidDFS2 (intx) {Vis2[x]=1; BOOLflag=false; for(inti=head[x];i!=-1; i=Nex[i]) { if(!vis2[to[i]]! ((x==stto[i]==en) | | (x==ento[i]==St))) {DFS2 (to[i]); Flag=true; dp[x][1]+=dp[to[i]][0]; dp[x][0]+=max (dp[to[i]][0],dp[to[i]][1]); }} dp[x][1]+=1ll*Fight[x]; if(!flag)

; i i) the { -dp[i][1] = val[0][i];Wuyi } the for(inti =0; i i) - { Wu for(intj =0; J j) - { About for(intK =2; K k) $ { -Dp[i][k] = max (Dp[i][k], dp[j][k-1] + val[j+1][i]); - intBBB =1; - } A } + } thecout 1][K] Endl; - $ the the /*cout the for (int i = 0; i the { - For (int j = 0; j in { the cout the } About cout the } the the cout + for (int i = 0; i -

The string appapt contains two words "Pat", where the first Pat is 2nd (P), 4th (a), 6th (t), and the second Pat is 3rd (p), 4th bit (a), 6th (t).Now given a string, how many pat can it form?Input Format:Input has only one row, contains a string, the length of not more than 105, only contains P, A, T three kinds of letters.output Format:Outputs the number of PAT contained in a given string in a row. Because the result may be larger, only the result of the remainder is output to 1000000007.Input

]) { theOther=e[j].to; the Break; - } -}ans=0; thememset (Vis,0,sizeof(VIS)); vis[root]=1; theSolve (Root,0, C[i]); theAns=max (ANS,TREEDP (Root,0, C[i])); theroot=other;deep[root]=0; -Other=i; thememset (Vis,0,sizeof(VIS)); vis[root]=1; thememset (tr,-1,sizeof(tr)); theSolve (Root,0, C[i]);94Ans=max (ANS,TREEDP (Root,0, C[i])); theend+=ans; the } the }98other=-1; Aboutmemset (Vis,0,sizeof(Vis)); - for(intI=1; iif(!Viss[i]) {101Root=i;solve

Title Description: Output the k-th prime number. Input: k≤10000 Output: The k-th prime number. Sample input: 37 Sample output: 517 Source: 2008 Shanghai Jiao Tong University computer Research Life Test real problem Answering: problem solving

The main topic: to four line segments, ask can form a rectangle?Title Analysis: First determine whether to form a quadrilateral, and then select an edge, see whether the other three edges in a parallel with him, two vertical.The code is as follows:# include　　Hihocoder #1040 (Judging whether it is a rectangle)

As easy as A+bTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 43019 Accepted Submission (s): 18379Problem Descriptionthese days, I am thinking on a question, how can I get a problem as easy as a+b? It is fairly difficulty to do such a thing. Of course, I got it after many waking nights.Give you some integers, your task was to sort these number ascending (ascending).Should know how easy the problem is now!Good luck!Inputinput contains multiple

Label: style blog Io color ar SP for div on #include Brute force for small case Pat 1040 longest shortric string

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1040The idea of solving problems: data is not big, can be directly bubble sort#include 　　Hangzhou Electric 1040 as easy as a+b "sort"