ccna training guide

Now do [email protected]@ will create a new 100 rows and add data to 103. Visual selection: V,V, Ahead, we saw theExamples of V and v. Once you've been chosen, you can do the following: j→ to connect all the rows (into one line) >→ indent around =→ Auto-indent Add something after all the selected lines: Select related rows (use j or /pattern or $ to the end of the lin

1. UVa 11300My Code:#include #include#includeusing namespacestd;Long Longc[1000010], M, A;intMain () {intN; while(~SCANF ("%d", N)) {c[0] =0; for(inti =1; I ) {scanf ("%lld", a); C[i]= c[i-1]+A; } M= c[n]/N; for(inti =1; I M; Sort (c, C+N); Long LongAns =0, x = c[n/2]; for(inti =0; I C[i]); coutEndl; }}View CodeAlgorithmic Competition Introduction Classic-Training guide

problem is how to use this method of describing state to further get the state transfer equation, easy to see, for Dp[1][n], initiator A has a series of methods, we enumerate this series (left K∈[1,n-1], right end take k∈[1,n]), and then get a score, Maintaining a maximum can be done to meet the constraints of the game (two people take the strategy to make their scores as high as possible, and two people are smart enough).It should be noted here that in min{} The first item in K is taken from i

calculation formula.where s = {a,b,c,d}.The reference code is as follows.#include #include#includeusing namespacestd;Const intMoD =1000007;Const intMAXN = the;intMain () {intC[MAXN][MAXN]; memset (C,0,sizeof(C)); for(inti =0; I ) {c[i][0] = C[i][i] =1; for(intj =1; J ) C[i][j]= (c[i-1][j-1] + c[i-1][J])%MoD; } //printf ("%d", c[15][14]); intT; scanf ("%d",t); intIcase =1; while(t--) { intsum =0; intN, M, K; scanf ("%d%d%d",n,m,k); for(inti =0; I -; i++) {

. A (n,m) = MN (m-1)(2) Vertical line: First put in a queen, there are NM methods, then another queen has n-1 method. i.e. B (n,m) = NM (n-1)(3) Slash: We first assume a size relationship--m>n, because this will limit the number of slashes we get, then based on this hypothesis, we will get the following diagonal length sequence:1,2,3,4...n-2,n-1,n,n,n...n,n,n (m-n+1 N) ... n-1,n-2,4,3,2,1, also take into account that there are two groups of such diagonal.So we are easy to know, C (n,m) = 2[2∑i (

structedge{int from, To,cap,flow;};structdinic{intn,m,s,t; Vectoredges; Vectorint>G[MAXN]; BOOLVIS[MAXN]; intD[MAXN]; intCUR[MAXN]; voidInitintN) {REP (I,1, N) G[i].clear (); Edges.clear (); } voidAddedge (int from,intTo,intcap) {Edges.push_back (Edge) { from, To,cap,0}); Edges.push_back (Edge) {to, from,0,0}); M=edges.size (); g[ from].push_back (M-2); G[to].push_back (M-1); } BOOLBFs () {MS0 (VIS); Queueint>Q; Q.push (s); D[s]=0; Vis[s]=1; while(!Q.empty ()) {

, distancetosegment (P, A, B)); Max= Max (max, Length (P-A)); Max= Max (max, Length (P-B));}intMain () {scanf ("%d",T); for(intKase =1; Kase ) {scanf ("%d%d",a,B); for(inti =0; i "%d%d",p[i].x,p[i].y); for(inti =0; i "%d%d",q[i].x,q[i].y); DoubleLenA =0, LenB =0; for(inti =0; I 1; i++) LenA + = Length (p[i+1] -P[i]); for(inti =0; I 1; i++) LenB + = Length (q[i+1] -Q[i]); intSa =0, Sb =0; Point Pa= p[0], Pb = q[0]; Min= 1e9, Max =-1e9; while(Sa 1 Sb 1) { DoubleL

remainder operation in number theory, involving the departure should be based on the inverse, and the size of the MoD and s[i] of the value, there is gcd (mod, s[i]), we can also get s[i] in about the MoD must exist in inverse element.Summing up, this topic is still a very good topic, it involves the basic principle of combinatorial counting, recursion, collation of mathematical formulas, basic tree-shaped data structure and the inverse of number theory, is a very comprehensive topic. Here the

Q: Matrix chain multiplicationStudents familiar with linear algebra know that matrix multiplication axb is defined when matrix A is m x N and Matrix B is n x p, and its operand is MNP.So now given n matrices, using array p[] (length n) to record the ranks of each matrix, then arrange a matrix multiplication allocation scheme, so that the global operation of the least amount.Analysis: It is obvious that this is a DP problem based on the linear interval, in fact, the DP problem is a big difficulty

++] = Centre_circletangenttwononparallellinewithradius (P1, v1, p2, v2, r); sol [ans++] = Centre_circletangenttwononparallellinewithradius (P1, V1 *-1, P2, v2, R); sol[ans++] = Centre_circletangenttwononparallellinewithradius (p1, v1, p2, v2 *-1, R); sol[ans++] = Centre_circletangenttwononparallellinewithradius (P1, V1 *-1, p2, V2 *-1, R); return ans;} A set of circles with two absent circles, three cases int Circletangenttotwodisjointcircleswithradius (Circle C1, Circle C2, double R, point *sol

structedge{int from, To,cap,flow,cost;};structmcmf{intn,m,s,t; Vectoredges; Vectorint>G[MAXN]; intINQ[MAXN]; intD[MAXN]; intP[MAXN]; intA[MAXN]; voidInitintN) { This->n=N; REP (i,1, N) G[i].clear (); Edges.clear (); } voidAddedge (int from,intTo,intCapintCost ) {Edges.push_back (Edge) { from, To,cap,0, cost}); Edges.push_back (Edge) {to, from,0,0,-Cost }); M=edges.size (); g[ from].push_back (M-2); G[to].push_back (M-1); } BOOLBellman_ford (intSintTintflow,intCost ) {REP (i,1,

) recursion is defined as follows: Make w = UNV, where n is the largest element of W , and you and v are shorter sequences; again S(w) = S (u) S (v) n, where S is the unit element for all the series with one element. C n represents the number of non-intersecting partitions for the collection {1, ..., n}.Then, Cn is never greater than the nth Bell number. C N also represents the number of non-intersecting divisions of the collection {1, ..., 2n}, where each paragraph has a length

, otherwise a fixed random value is generated ... Hey, Daddy. for(intI=1; i) {scanf ("%lld",A[i]); Tot+=A[i]; } Long Longave=tot/N; c[0]=0; for(intI=1; i) {C[i]=c[i-1]+ave-a[i+1];//cout} sort (C,c+N); Long Longmid=c[n/2]; Long Longans=0; for(intI=0; iC[i]); printf ("%lld\n", ans); } return 0;}I found such a problem when I was reading a book. What is the difference between fabs and abs? What library do you need? It seems that ABS does not require additional libraries.Libraries