intap

1001, HDU 5327 OlympiadPlay a table check-in question. The degree of opening 4minFB ...There should be a more optimized method, in order to hand speed without tle on the line ... = =1002, HDU 5328 zzx and permutationsAP: Linear sequenceGP: equal-ratio sequenceThe length of the maximum sequence of AP and GP is calculated by the ruler, then the maximal value can be obtained.It is important to note that the equivalence is best used in a proportional way rather than in division, because the sample,

]-(j-k) *ap[i],f[i][j]) F[i][j]=max (F[i-w-1][k]+k*ap[i])-j*ap[i] Make g[i-w-1][k]=f[i-w-1][k]+k*ap[i]→ F[i][j]=max (g[i-w-1][k])-j*ap[i] 3, from i-w-1 days sell shares: F[i][j]=max (f[i-w-1][k]+ (k-j) *bp[i],f[i][j]) F[i][j]=max (F[i-w-1][k]+k*bp[i])-j*bp[i]. Make G ' [i-w-1][k]=f[i-w-1][k]+k*bp[i]→ F[i][j]=max (g ' [i-w-1][k])-j*bp[i] Because for G[i-w-1][j] and G ' [i-w-1][j] are satisfied: if J1>j2 and G[i-w-1][j1]>g[i-w-1][j2], you do not have to keep g[i-w-1][j2]. S

=sc.nextint (); Sc.nextline ();if(n==0){return; } String str=NULL;CharA[] =New Char[(n+1)/2];CharB[] =New Char[(n+1)/2];intap=0;intbp=0;intanum=0;intbnum=0; for(intI=0; iBooleanIsNo =false;if(Str.endswith ("No Good")) {isno=true; }if(IsNo) {if(i%2==0) {a[ap++]=' X '; }Else{b[bp++]=' X '; } }Else{if(i%2==0) {a[ap++]=' O '; anum++; }Else{b[bp++]=' O '; bnum++; } } } for(intI

=GetChar (); while(ch'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} while(ch>='0'ch'9') {x=x*Ten+ch-'0'; ch=GetChar ();} returnx*F;}intMain () {n=read (); P=read (), w=read (); intAP,BP, as, BS; intans=-INF; Memset (DP,-inf,sizeof(DP)); for(intI=1; i) {AP=read (), Bp=read (), as=read (), bs=read (); for(intj=0; j as; j + +) dp[i][j]=-ap*J; for(intj=0; j1][j]); intk=i-w-1; if(k>=0) { intL=0, r=0; for(intj=0; j) {

v;} as[1007*555];BOOL operatorConstLinea,ConstLINEAMP;B) {returna.vB.V;}intap=0, ans=0;intes[1000007],enx[1000007],e0[1007],ep=2, ds[1007],dp=0, t,tk[1007];intDfsintW) { intC=1; TK[W]=T; for(intI=e0[w];i;i=Enx[i]) { intu=Es[i]; if(tk[u]!=t) c+=dfs (U); } returnC;}voidAdde (intAintb) { if(!e0[a]) ds[dp++]=A; if(!e0[b]) ds[dp++]=b; ES[EP]=b;enx[ep]=e0[a];e0[a]=ep++; ES[EP]=a;enx[ep]=e0[b];e0[b]=ep++;}intMain () {scanf ("%d",N); for(in

maintenance of the subtree" and for most of the DFS sequence maintenance sequence of the problem, and the tree array is quite good combination of the problem is almost naked, open an array that represents the current node state can be1#include 2#include 3#include string.h>4 intap[100010],bit[100010];5 intbg[100010],ed[100010],cnt=0;6 intN;7typedefstruct{8 intto,nxt;9 }edge;TenEdge gra[200010]; One inthead[100010],num=0; A intAddintfrmintTo ) { -g