cisco 1006

This article is reprinted, Source Address: http://blog.csdn.net/dongfengkuayue/article/details/6461298 poj 1006 biorhythms Time limit:1000 ms Memory limit:10000 K Total submissions:78980 Accepted:23740 DescriptionSome people believe that there are three cycles in a person's life that start the day he or she is born. these three cycles are the physical, emotional, and intellectual cycles, and they have periods of l

VIP cannot be started normallyDescription: Our environment is 2 node RAC, and node 1 is down due to physical faults.In this case, I want to start the VIP of Node 1 from node 2 so that the single node is transparent to the user program. [Oracle @ UNID02 ~] $ Crs_start ora. unid01.vipAttempting to start 'ora. unid01.vip 'on member 'unid02'Start of 'ora. unid01.vip 'on member 'unid02' failed.CRS-1006: No more members to consider CRS-0215: cocould not sta

first, the requirements http://poj.org/problem?id=1006 biorhythms Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 124113 Accepted: 39154 DescriptionSome people believe that there be three cycles in a person's life that's start the day he or she's born. These three cycles is the physical, emotional, and intellectual cycles, and they have periods of lengths, D, and Ays, respectiv

Time: 2016-05-09-21:12:54 Title Number: [2016-05-09][51nod][1006 longest common sub-sequence LCS] Main topic: [2016-05-09][51nod][1006 longest common sub-sequence lcs].md Analysis: Dynamic Planning DP[I][J] Represents the length of the longest common subsequence of string A in the first position, string B at position J DP[I][J] = dp[i-1][j-1] + 1 if a[i] = = A[j] else dp[i

1006. Summation Game DescriptionThere is a row of stone keyboards on the stone pillars, with an integer on each key. Please select two keys on the keyboard to make the keys between the two keys and the number and maximum. If this is the largest and not positive, then output "Game over".Input FormatLine 1th: Number of keys N.2nd.. N+1: The numeric integer on the key ai. −100≤ a I≤100 For 70% of data, 2n ≤1 ,000 For 100% of data,2≤n≤1, O

); - } - } - } in intMain () - { to intT; +scanf"%d",T); - while(t--) the { *scanf"%d%d",n,m); $ for(intI=1; i) G[i]. Clear ();Panax NotoginsengMemsetinch,0,sizeof(inch)); -Memset (Val,0,sizeof(Val)); the while(m--) + { A intu,v; thescanf"%d%d",u,v); + G[u].push_back (v); - inch[v]++; $ } $ solve (); - intminn=100000+5; -ll ans=0; the for(intI=1; i) - {Wuyiminn=min (minn,val[i]); theans+=(

classes.User Management class: Used to manage user data, verify user name and password through database after user login, and read user's details from database after authentication. While maintaining user data in the program, the user data is saved to the database when the user game ends and exits the game.Server Framework class: Used to manage the game hall data, including a list of some data.Database classes: Server-side for network games when processing large amounts of customer data, using

Physiological cycle POJ 1006, physiological cycle poj1006 Time Limit:1000 MS Memory Limit:10000 K Total Submissions:138101 Accepted:44225 DescriptionThere are three physiological cycles in life: physical strength, emotional strength, and mental strength. The cycle is 23 days, 28 days, and 33 days. One day in each cycle is the peak. On the peak day, people will do well in the corresponding aspects. For example, at

voidMakeheight (intN) - { About intI, j, k =0; $ for(i =0; I i; - //the nth character ranking is 0,sa[rank[i]-1] out of bounds, so don't count - for(i =0; ik) - for(K. k--:0, j = Sa[rank[i]-1]; R[i + K] = = R[j + K]; k++); A } + the CharStr[n]; -STD::stringAns1,ans2; $ intidx1,idx2; the intMain () the { the intt,n,m; thescanf"%d",t); - while(t--) in { theans1 = Ans2 =""; thescanf"%d",n); Aboutscanf"%s", str); the for(intI=0; ii) theR[i] = str[i]-'a'+1;

According to the statistics system of hacked sites in China, from September to 3rd, the total number of submitted illegal websites in China was 1006, up 68.79% from last week. The following is the specific situation: 650) this. width = 650; "src =" http://www.bkjia.com/uploads/allimg/131227/0941512264-0.jpg "alt =" 20130922053913832.jpg"/>Figure 1) submitted by the hacked website in week 5, 3rd from September) As shown in figure 1, a total of 517 subm

) - { in Long LongTemp,ret; thetemp = (all-size[c[x])%MoD; the if(! is[x]) { About if(have[c[x]].size () = =1) theRET =temp; the Else theret = (Temp+size[c[x]]*ni (a[x))%MoD; + } - ElseRET = temp+F[x]; the return(ret%mod+mod)%MoD;Bayi } the the intMain () - { - inttest,i,j,x,y; the for(Cin >> test; test--; ) { theCIN >> N >>m; the for(i =1; i) thescanf"%i64d", a[i]); - the for(i =1; i) theHead[i] =0; thep =0;94 the

Title Link: The Magical kingdomAn essay question ... Magic string diagram, Magical mcs ...Feel like I have nothing more to say, here's a brief introduction to MCS:We give each point a weighted value, from the back to the next step to determine the perfect elimination sequence of points, each time the choice of the maximum value of a point (the same words randomly selected) placed in the current perfect elimination sequence position, and then the adjacent to a bit of weight added \ (1\). A perfec

dead. */#include#definell Long Long#defineINF 0X3FFFFFFFFFFFFFFF#defineN 22using namespacestd;stringOp;ll Right (strings) {ll e=0; intn=s.size (); for(intI=2; i) e=e*Ten+s[i]-'0'; //coutll ans= (s[0]-'0') * (s[1]-'0')/e; returnans;} ll left (stringS//the left part of the minus sign{ll Cur1=0, cur2=0; intn=s.size (); for(intI=1; i) Cur1=cur1*Ten+ (s[i]-'0'); Cur1+=s[0]-'0'; for(intI=0; i1; i++) Cur2=cur2*Ten+ (s[i]-'0'); CUR2+=s[n-1]-'0'; //cout returnMax (CUR1,CUR2);} ll solve (strings) {

Topic linksDescriptionGiven two string a B, the longest common subsequence of A and B (the subsequence is not required to be contiguous).For example, two strings: ABCICBA Abdkscab AB is a sub-sequence of two strings, and ABC is also, ABCA, where ABCA is the longest subsequence of the two strings.InputLine 1th: String ALine 2nd: string b(A, B's length OutputOutput the longest sub-sequence, if there are multiple, randomly output 1.Input exampleAbcicbaAbdkscabOutput ExampleABCAThe code is as follow

back and forth on both paths.Input and Output Sample input example # #:3 30 3 92 8 55 7 0Sample # # of output:34DescriptionLimit30% of the data meet: 1100% of the data meet: 1NOIP 2008 Raising the third question of the groupDP Category: Checkerboard type DP#include #includeusing namespacestd;#defineN 51intN,m,s[n][n],f[n][n][n][n];intMain () {scanf ("%d%d",n,m); for(intI=1; i){ for(intj=1; j) {scanf ("%d",S[i][j]); } } for(intI=1; i){ for(intj=1; j){ for(intk

forth on both paths.Input and Output Sample input example # #:3 30 3 92 8 55 7 0Sample # # of output:34DescriptionLimit30% of the data meet: 1100% of the data meet: 1NOIP 2008 Raising the third question of the group/*two-way transmission of the note, can be seen from the starting point has two pieces of paper together F[i][j][k][l] said from the beginning to (I,J) and (K,l) the greatest affection of the note can be from the top, or the top and left, or left and top, or all from the left when al

#includestring.h>#include#include#include#includeusing namespacestd;Const intn=1e5+Ten;Const intinf=0x3f3f3f3f;Const intmod=1e9+7; typedefLong LongLL;intA[n], flag;voidIncrease (intN) { inti, NUM, idex; Num= Flag =0; for(i =2; I ) { if(A[i] 1]) {num++; Idex= i;///Mark subscript that does not satisfy an element } } if(num = =0) flag =1; if(num = =1) { if(Idex = = N | | idex = =2|| a[idex-2] 1] 1]) Flag=1;///If this does not satisfy the left and right side of

occurs in 10789 days.SourceEast Central North America 1999TranslatorBeijing University Program Design Internship, Xie Di1#include 2#include 3#include 4#include 5#include 6#include 7 using namespacestd;8 inta[Ten],m[Ten],tot,now,m;9InlinevoidEXGCD (intAintBintd,intx,inty) {Ten if(b==0){ Onex=1; y=0; D=A; A return ; - } -EXGCD (B,A%B,D,Y,X); Y-=x* (A/b); the } -InlineintChina (intR) { -m=1; - intmi,d,x0=0, y0=0, ans=0; + for(intI=1; iM[i]; - for(intI=1; i){ +mi=m/M[i

occurs in 1234 days.Use the plural form "days" even if the answer is 1.Sample Input0 0 0 00 0 0 1005 20 34 3254 5 6 7283 102 23 320203 301 203 40-1-1-1-1Sample OutputCase 1:the Next triple peak occurs in 21252 days. Case 2:the Next triple peak occurs in 21152 days. Case 3:the Next triple peak occurs in 19575 days. Case 4:the Next triple peak occurs in 16994 days. Case 5:the Next triple peak occurs in 8910 days. Case 6:the Next triple peak occurs in 10789 days.SourceEast Central North America 19

Test instructions: number n, is (n+d)%23==p, (N+d)%28==e, (n+d)%33=i;Reprint please indicate source: http://www.cnblogs.com/dashuzhilin/;Idea: Chinese remainder theorem. Using the additive of congruence, the (n+d) is split into three number a,b,c,Make a%23==p,a%28==0,a%33==0;Make b%23==0,b%28==e,b%33==0;Make c%23==0,c%28==0,c%33==i;Then (n+d) = = (A+b+c) +LCM (23,28,33) *t;So, we can do the optimization, initially, the p,e,i are 1, namely:Make the A%23==1,a%28==0,a%33==0;a a multiple of 28 and 3