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In an application for study in the United States, a lot of people will be for some professional choice and puzzled, mainly on the choice of professional information is not clear enough, especially in recent years the rise of professional, due to professional reference information is not complete, so it is difficult to choose, the following is a look at the U.S. financial science PhD and Economics of some differences, I hope I can give you some help. The American PhD in economics is two grades wo

Code:#include POJ 1236 Network of schools (strongly connected component template problem)

[v]); if(Low[v] >Dfn[u]) {Bridge[v]=1; Brinum++; } } Else if(v! = PA) Low[u] =min (low[u],dfn[v]); }}voidLcaintAintb) { while(Deep[a] >Deep[b]) { if(Bridge[a]) {Bridge[a]=0; Brinum--; } A=Fa[a]; } while(Deep[b] >Deep[a]) { if(Bridge[b]) {Bridge[b]=0; Brinum--; } b=Fa[b]; } while(A! =b) {if(Bridge[a]) {Bridge[a]=0; Brinum--; } if(Bridge[b]) {Bridge[b]=0; Brinum--; } A=Fa[a]; b=Fa[b]; }}intMain () {intN,m,q,a,b,ca =0; while(~SCANF ("%d%d",n,m)) {if(!

++] =u; Instack[u]=true; intLen =g[u].size (), V; for(intI=0; i) {v=G[u][i]; if( !Low[v]) {Tarjan (v); Low[u]=min (Low[u], low[v]); } elseif (Instack[v]) {Low[u]=min (Low[u], dfn[v]); } } if(Low[u] = =Dfn[u]) { Do{v= stack[--top]; BELONG[V]=CNT; INSTACK[V]=false; } while(U! =v); CNT++; }}voidsolve () {intINDEGREE[MAXN] = {0}; intOUTDEGREE[MAXN] = {0}; intin =0, out =0; for(intI=1; i) { if(!Low[i]) Tarjan (i); } for(intI=1; i) { intLen =g[i].s

], which represents the problem (the first I-character of a s1. The state (true or false) of the S2 's first J-character, S3 's first i+j character). Then C[s1.length ()][s2.length ()] is the state of the target problem. The initial problem state c[0][0]=true. Status recursion relationship: C[i][j] = ((s1[i-1] = = s3[i+j-1]) c[i-1][j]) | | ((s2[j-1] = = s3[i+j-1]) c[i][j-1]) Code: Class Solution {public : bool Isinterleave (string s1, String s2, string s3) { size_t len1, len2;

; the} while(u!=v); - } Wu } - intTarjan () About { $sum=0;d f=0; sttop=0; - for(intI=1; i) - if(!Dfn[i]) - Tarjan_dfs (i); A for(intI=1; i) +first[i]=0;//empty the original edge the for(intI=1; i) - if(Tar[e[i]. from]!=tar[e[i].to]) $ { theE[i]. from=tar[e[i]. from]; thee[i].to=tar[e[i].to]; theE[i].next=first[e[i]. from]; theFirst[e[i]. from]=i; -num[e[i].to]++; in } the returnsum; the } About voidDfsintk) the { the if(First[k]) the for(i

encountering a major accident in the tunnel, escape appears to be especially critical, if the tunnel distance is not long, ensure the safety of the premise to escape the tunnel along the roadside best. If the tunnel is long or the road has been blocked, you can evacuate using the personnel escape route.Shutter Doors escape: The fire door of the car lanes adopt the form of shutter doors, under normal circumstances in the closed state, in case of emergency can be controlled by three ways: one is

Water problem, the direct column equation then solves can. Assuming a growth rate of one Earth's resources V, then the largest population is the rate at which these populations consume resources at a speed of V is the number. So: A*v-a*x=b*v-b*y can solve the answer.#include #includeusingnamespace std; Double int main () { scanf ("%lf%lf%lf%lf", x,a ,y,b); printf ("%.2f", (x*a-y*b)/(A-b));}Estimation of population carrying capacity of Olympiad Earth in primary

position Overtaking lane change accidentFour shots: Front, rear, two-car relative marking position, collision position Island Type AccidentsTake a picture: Take a car at the location of a car that goes into or out of the roundabout. Through the above introduction, I believe we have a certain understanding of how to take photos after an accident, if the future encounter similar accidents, I hope that in a timely manner after taking evidence, the vehicle moved to not obstruct th

found and slowed down in time. Try to avoid stopping on the slope in order to avoid the situation of leaving the car. Try to get off from the right to avoid collisions with the car.A thorough inspection of the vehicle after the return of the expeditionAfter the trip back people tired car also tired, so when you have a good rest after the proposed to check the situation of the vehicle, in fact, not complicated, to see if the tires have scars, oil is not missing and so on. Of course, if you have

Simulate with two stacks and save a timestamp for each point. The time stamp of the merge is recorded at each merge mcnt and the Topa and TOPB at this time do TA, TB.Every time the pop, if the stack top timestamp is greater than mcnt, then normal pop, or in two stacks TA and TB under the maximum time stamp and not yet pop off. Then use the bj[time stamp] to mark the pop already. #include WA has several hair, the result is "if (a[topa].id>mcnt)" Write is >=, why equal not, because POPC

Topic linksLook at the left parenthesis as a right parenthesis as B and push it. It was written a long time ago and pushed to the last discovery is a regular sequence.#include using namespaceStd;typedefLong Longll;intMain () {ll n; while(SCANF ("%lld", n)! =EOF) {ll CNT=0, sum=0; for(intI=1;; i++) { if(sum>=n) Break; CNT++; Sum=sum+CNT; //printf ("%d:%d\n", i,sum);} printf (")"); ll a=sum-N; LL b=cnt-A; for(intI=1; i) { if(b==i) printf ("("); Elseprintf (")

Title Address: HDU 5328Test instructions: In a sequence of length n to remove the number of consecutive k, so that the number of k composition arithmetic progression or geometric series, ask such k maximum can be how much.Ps: Pay attention to double, because the geometric series divide may be decimal.#include #include #include #include #include #include #include #include #include #include #include #pragma COMMENT (linker, "/stack:102400000,102400000")using namespace STD;typedef Long LongLL;Const

(intj=1; j7; j + +) $dp[i][j]=dp[i-1][j]; -dp[i][ans]++; - } the } - intMain ()Wuyi { the intt,l,r,aa[Ten],ma; - getprime (); Wu Getcot (); - Gao (); Aboutscanf"%d",T); $ while(t--) - { -Ma=1; -scanf"%d%d",l,R); A for(intI=1; i7; i++) + { theaa[i]=dp[r][i]-dp[l-1][i]; - if(aa[i]>=2i>Ma) $Ma=i; the } the if(aa[6]>0aa[3]>0) theMa=max (MA,3); the Else if(aa[6]>0aa[2]>0|| aa[4]>0aa[2]>0) -Ma=max (MA,2); inprintf"%d\n", MA); the

); while (T--) {scanf("%d%d",L,R); int Record[8];Memset(Record,0 ,sizeof( Record)); for (I=1;I7;I++) {Record[I]=Leap[R][I]-Leap[L-1][I]; } int Ans=1; for (I=1;I7;I + +) { if( Record[I]==0 ) continue; if (Record[I]>1)Ans=Max(Ans,I); for (J=I+1;J7;J + +) { if( Record[J]==0 ) continue; else if (I==2J==4 ) { if( Ans2)Ans=2; } Else if( I==2J==6 ) { if( Ans2)Ans=2; } Else if( I==4J==6 ) { if( Ans2)Ans=2; } Else if( I==3J==6 ) { if( Ans3)Ans=3; } } }Printf("%d\n",Ans);

Original title Link http://acm.hdu.edu.cn/showproblem.php?pid=5410Because he gets an AI candy for every item he buys, he gets bi sugar for every item he buys. There are two values: So we should first use a 01 backpack to unify the two values ... Because each item in the 01 backpack can only be taken once .... Then we use multiple backpacks to decide which items we should buy more of the most candy .... ==#include #includestring.h>#include#includeusing namespacestd;Const intmaxn=1010; __int64 dp[

Topic Link: Click to open the linkThe main idea: given the initial value of a n*n matrix, and the final value, there is now M operation L I J, the value of column I is reset to J,h i J, the value of line I is reset to J. Ask how the M operation should be done and the matrix can be changed.Looking forward from the final value of the scheme, each time to find the row and the remaining colors in the column are all the same, see if there is no use of the operation can complete it, if there is record

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5316Test Instructions: give you n points, m operations, each operation has 3 integers t,a,b,t represents the operation type, when t=1 the value of a point is changed to B; When T=0, the largest sub-sequence between the query interval a, B, and the original subscript parity of the adjacent elements in this subsequence are different. Ideas: the difficulty of this problem is in the query, the rest are templates, and according to the query, you j

Topic Link: Click to open the linkDefinition f (i) for the type of prime number comprising I, the maximum value of gcd (f (i), F (j)) (L The prime number sieve 10^6 in the prime, the F (i) of each count, can be found F (i) subtract directly from the calculation interval and judge the maximum GCD#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. HDU5317 (more than 2015 schools 3)--RGCD

Title Address: HDU 5414Test instructions: Ask to determine whether the string s can be added several characters to get the string tIdea: This problem looks complex, in fact, a careful analysis, successful conversion contains only two cases. The first is that because the new character required to be inserted differs from the character C in front of it, if there is x continuous C in T, then there must be x consecutive C in S, and the second is a discontinuous substring of t that must be s.#include