intrr

Constaconst_rvalue_reference2=const_lvalue;//error Constaconst_rvalue_reference3=rvalue ();//ok Constaconst_rvalue_reference4=const_rvalue ();//ok Rule five: Exception for function type Voidfun () {} Typedefdecltype (Fun) fun;//typedefvoidfun (); Funlvalue_reference_to_fun=fun;//ok Constfunconst_lvalue_reference_to_fun=fun;//ok Funrvalue_reference_to_fun=fun;//ok Constfunconst_rvalue_reference_to_fun=fun;//ok "description":(1) Some support right reference but the lower version of

+ - using namespacestd; + A Const intMAXN =100010; at intN, S; - intX[MAXN]; - intSUM[MAXN]; - - BOOLOkintmm) { - for(inti =1; I 1; i++) { in //printf ("%d%d\n", i-1, i+mm-1); - if(sum[i+mm-1]-sum[i-1] >= s)return 1; to } + return 0; - } the * intMain () { $ //freopen ("in", "R", stdin);Panax Notoginseng intT; -scanf"%d", T); the while(t--) { +memset (SUM,0,sizeof(sum)); Ascanf"%d%d", n, s); the for(inti =1; I ) { +scanf"%d", x[i]); -Sum[i

Level 0: Sacred table (∞)1#include 2 intMain () {3 intarr[ One] = {0,1,0,0,2,Ten,4, +, the,352,724};4 intN;5 while(~SCANF ("%d", n) N) {6printf"%d\n", Arr[n]);7 }8}Level 1: One-dimensional violence backtracking (13)1 Const intMAXN = -;2 intG[MAXN];//G[c] = r3 intN, CNT;4 5 BOOLOkintcc) {6 for(intRR =0; RR ) {7 if(G[RR] = = G[CC] | | ABS (RR-CC) = = ABS (g[rr]-G[CC])) {8 return false;9 }Ten } One

can be divisible by 4After 1582 years (including 1582), the GC judges the criteria for a leap year (any one of the following):(1) can be divisible by 4, but not divisible by 100;(2) can be divisible by 400.For historical reasons, the GC prescribes 1700 as a leap year unconditionallyDue to historical reasons, the GC stipulates that September 3, 1752 ~ 13th a total of 11 days does not exist, that is, September 1752 only 19 daysGC has 7 days in a week, sorted for Sun,mon,tue,wed,thu,fri,sat, and n

; }intMid = (L + r)/2; Build (Lson (U), L, mid); Build (Rson (U), Mid +1, R); Pushup (u);}voidModify (intUintXintV) {if(Lc[u] = = x Rc[u] = = x) {A[x] = V;return; }intMID = (Lc[u] + rc[u])/2;if(x ElseModify (Rson (U), X, V); Pushup (u);}intQuery (intUintLintR) {if(l returnS[u];intMID = (Lc[u] + rc[u])/2, ret;if(R Else if(L > Mid) ret = query (Rson (U), L, R);Else{intll = query (Lson (U), L, R);intrr = Query (Rson (U), L, R);intA = min (R[lson (u)]

].lm+=tree[rt1|1].LM; if(tree[rt].rm==tree[rt1|1].len) tree[rt].rm+=tree[rt1].rm; TREE[RT].NM=max (tree[rt].nm,tree[rt1].rm+tree[rt1|1].lm); }}voidBuildintLintRintRT) {TREE[RT].L=M; TREE[RT].R=R; Tree[rt].len=r-l+1; if(l==r) {TREE[RT].LM=tree[rt].rm=tree[rt].nm=1; Tree[rt].ln=tree[rt].rn=Num[l]; return; } intMid= (l+r) >>1; Build (Lson); Build (Rson); Pushup (RT);}voidUpdateintPosintXintLintRintRT) { if(l==posr==POS) {Tree[rt].ln=tree[rt].rn=x; return; } intMid= (l+r) >>1; if(posm

; - }; - - intOkintXintYintv) - { in inti,j; -i =x; to for(j =1; J 9; j + +){ + if(Ans[i][j] = = v)return 0; - } thej =y; * for(i =1; I 9; i++){ $ if(Ans[i][j] = = v)return 0;Panax Notoginseng } - intrr = (X-1)/3*3; the intCC = (y1)/3*3; + for(i = RR +1; I 3; i++ ){ A for(j = cc +1; J 3; j + + ){ the if(Ans[i][j] = = v)return 0; + } - } $ return 1; $ } - - void

; } intm = (L + r) >>1; if(P m) Update (p, x, Lson); ElseUpdate (P, x, Rson); Pushup (L, R, RT);}intQueryintLintRintLintRintRT) { if(l r) {returnlen[rt].m; } intm = (L + r) >>1; if(R m)returnquery (L, R, Lson); Else if(L >m)returnquery (L, R, Rson); Else { intLL =query (L, R, Lson); intRR =query (L, R, Rson); intRET =Max (LL, RR); if(flag) {ll= Min (len[rt 1].R, M-l +1);//cannot exceed the length of the query intervalrr = min (len[rt 1

] = = 0) { -bb[cnt++] =i; - } - } - intSS = 1; in while(ss = = 1) { -K =in.nextint (); to intAns = 0; + if(k = = 0) { - Break; the}Else { * intsum = 0; $ intLL = 0;Panax Notoginseng intRR = 0; - while(ss = = 1) { the if(RR >CNT) { + Break; A } the

, obtained from the son nodetree[rt].len=tree[rt1].len+tree[rt1|1].len;}voidUpdata (intLintRintCintLintRintRT) { if(l>=lr>=r) {tree[rt].cnt+=C; Getlen (RT,L,R); return ; } intM= (L+R)/2; if(m>=L) Updata (L,r,c,lson); if(r>m) updata (L,r,c,rson); Getlen (rt,l,r);}intMain () {intn,m,t,i,j,ff=0; DoubleX1,x2,y1,y2; while(SCANF ("%d", n)! =EOF) { if(!N) Break; M=0; for(i=0; i) {scanf ("%LF%LF%LF%LF",x1,y1,x2,y2); S[M].L=x1;s[m].r=x2;s[m].h=y1;s[m].f=1; //Lower Borders[m+1].l=x1;

)/2; if(m>=L) Updata (L,r,c,lson); if(r>m) updata (L,r,c,rson); Getlen (L,R,RT);}intFindDoubleValintXinty) { intL=x,r=y,m; while(lr) {m= (l+r)/2; if(mark[m]==val)returnm; Else if(mark[m]>val) r=m-1; Elsel=m+1; } return-1;}intMain () {intT,n,i; DoubleX1,x2,y1,y2; scanf ("%d",t); while(t--) {scanf ("%d",N); intm=0; for(i=0; i) {scanf ("%LF%LF%LF%LF",x1,y1,x2,y2); S[M].L=x1;s[m].r=x2;s[m].h=y1;s[m].f=1; mark[m++]=X1; S[M].L=x1;s[m].r=x2;s[m].h=y2;s[m].f=-1; mark[m++]=x2; } sort (S,s+m,cmp)

("%d", n); Rep (I,0, N-1){scanf("%d%d%d", AMP;POINT[I].X,AMP;POINT[I].Y,AMP;POINT[I].R); } sort (point,point+n,cmp); Tail =0; MEM (head,-1); Rep (I,0, N-1){intFlag =0; Rep (j,i+1, N-1){intrr = (LL) POINT[J].R * POINT[J].R;intdis = dist (i,j);if(RR > Dis) {flag =1; Add (j,i); Break; } }if(!flag) Add (n,i); }if(DFS (n)! =0)puts("Alice");Else puts("Bob"); }return 0;} Copyright NOTICE: This article for Bo Master original artic

(left) {} nodeoperator+=(Node B) {num+=B.num; Left=B.left; return* This; }}dp[ -][ About][1010];node DFS (intIintSumintLeftintLeintre) { if(i==0){ if(sum+left>=k)returnNode1,0); returnNode0, sum+Left ); } if(used[i][sum][left]!le!re)returnDp[i][sum][left]; intLl=le?lbit[i]:0; intRr=re?rbit[i]:9; Node A (0, left); for(intv=ll;vv) {a+=dfs (I-1,sum+v,a.left,le (V==LL),re (v==RR)); } if(!le!re) {Dp[i][sum][left]=A; Used[i][sum][left]=1;

(str[now]==0) str[now]=str[now-1]; Else{Change ( now+1); Str[now]=str[now-1]; }}voidQread (intx) {x=0; vn=GetChar (); while(vn'0'|| Vn>'9') vn=GetChar (); while(vn>='0'vn'9') {x=x*Ten+(int) (vn-'0'); vn=GetChar ();}}voidCheckintNowCharty) { intl=now-1, r=now,nl=0, nr=0, pd=0; while(1) { if(str[l]==ty) nl++,l--; if(str[r]==ty) nr++,r++; if(Str[l]!=tystr[r]!=ty) Break; } if(nl+nr+1>=3) { while(1) {NL=0, nr=0, ty=Str[l]; intLl=l,rr=R; intKol; while(1) {Kol=0; if(str[l

to be added.tree[rt].num-=2; }}voidUpdata (intLintRintCintLintRintRT) { if(lr) {tree[rt].cnt+=C; Getlen (L,R,RT); return ; } intM= (L+R)/2; if(m>=L) Updata (L,r,c,lson); if(r>m) updata (L,r,c,rson); Getlen (L,R,RT);}intFindintValintnum) { intl,r,m; L=0; r=num; while(lr) {m= (l+r)/2; if(val==Mark[m])returnm; Else if(val>Mark[m]) L=m+1; Elser=m-1; } return-1;}intMain () {intI,n,m,l,r; intX1,x2,y1,y2; while(SCANF ("%d", n)! =EOF) {m=0; L=999999; R=-999999; memset (s),0,sizeof(

#include #include#include#includeusing namespacestd;intrr,cc;Chars[ -][ -];intnum[ -][ -];BOOLBorderintXinty) { if(x>=0x0AMP;AMP;Yreturn true; return false;}intmove[2][2]={{0,-1},{-1,0}};BOOLCheckintXinty) { if(s[x][y]=='*')return false; for(intI=0;i2;++i) { inttx=x+move[i][0]; intty=y+move[i][1]; if(!border (Tx,ty))return true; if(s[tx][ty]=='*')return true; } return false;}Charb[ $][ $];Charc[ $][ $];Chartemp[ $];intb[ $],c[ $];intM

", N, T/N, t ); Break; } Q. Push (T * 10 ); Q. Push (T * 10 + 1 ); } } Return0; } Method 4 source code: // Solution 4: divide the search space into the breadth of the category, so that the space is occupied by O (n) instead // Exponential level. The classification principle is to classify according to the remainder of model N. # DEFINE _ crt_secure_no_warnings 1 # Include # Include # Include # Include UsingNamespaceSTD; StructQnode { IntV, R; // v is value, R is Remai

temporary object, you need to:1. The referenced object will be destroyed2. The object has no other user2. Applicationfor Lvalue references, you cannot bind them to an expression that requires conversion, a literal constant or an expression that returns a right value. An rvalue reference can be bound to such an expression, but you cannot bind an rvalue reference directly to an lvalue. Functions that return non-reference types, together with arithmetic, relationships, bits, and post increment/dec

; for(i=1; ii; for(i=0; ik) for(k?k--:0, j=sa[rank[i]-1]; R[I+K]==R[J+K]; k++);}intMain () {intT,len,case=0;; Charch; scanf ("%d",t); while(t--) { case++; ll Ans2=0; Memset (R,0,sizeof(R)); scanf ("%c",ch); scanf ("%s", str); Len=strlen (str); intRr=Len; for(inti=len-1; i>=0; i--) { if(str[i]!=ch) {R[i]=RR; } Else{ll h= (rr-i-1); RR=i; R[i]=RR; Ans2+=h* (H +1)/2; }} ans2+=rr* (rr+1)/2; //coutDa (str,sa,len+1, the); Calhe

coordinates on the picture (lazy)When it's not the sameif out pixeluv)) { 2; 2 ; }Fourth step, set the color value of the corresponding position.I've used two different methods. Public voidDrawRect (intXinty, color color, Vector2 drawrect) { intStartX = X-(int) (Drawrect.x)/2; intStarty = y-(int) (DRAWRECT.Y)/2; for(inti = startx; I int) (Drawrect.x); i++) { for(intj = Starty; J int) (DRAWRECT.Y); J + +) {t2d.