function gcd(a,b:integer):integer;
begin 
if b=0 then gcd:=a
else gcd:=gcd (b,a mod b);
end ; 

function lcm(a,b:integer):integer;
begin
if a<b then swap(a,b);
lcm:=a;
while lcm mod b>0 do inc(lcm,a);
end; 

function prime (n: integer): Boolean;
var I: integer;
begin
for I:=2 to trunc(sqrt(n)) do
if n mod I=0 then begin 
prime:=false; exit;
end;
prime:=true;
end; 

procedure getprime;
var 
i,j:longint;
p:array[1..50000] of boolean;
begin
fillchar(p,sizeof(p),true);
p[1]:=false;
i:=2;
while i<50000 do begin
if p[i] then begin
j:=i*2;
while j<50000 do begin
p[j]:=false;
inc(j,i);
end;
end;
inc(i);
end;
l:=0;
for i:=1 to 50000 do
if p[i] then begin
inc(l);pr[l]:=i;
end;
end;{getprime}

function prime(x:longint):integer;
var i:integer;
begin
prime:=false;
for i:=1 to l do
if pr[i]>=x then break
else if x mod pr[i]=0 then exit;
prime:=true;
end;{prime} 

procedure prim(v0:integer);
var
lowcost,closest:array[1..maxn] of integer;
i,j,k,min:integer;
begin
for i:=1 to n do begin
lowcost[i]:=cost[v0,i];
closest[i]:=v0;
end;
for i:=1 to n-1 do begin
{尋找離生成樹最近的未加入頂點(diǎn)k}
min:=maxlongint;
for j:=1 to n do
if (lowcost[j]<min) and (lowcost[j]<>0) then begin
min:=lowcost[j];
k:=j;
end;
lowcost[k]:=0; {將頂點(diǎn)k加入生成樹}
{生成樹中增加一條新的邊k到closest[k]}
{修正各點(diǎn)的lowcost和closest值}
for j:=1 to n do
if cost[k,j]<lwocost[j] then begin
lowcost[j]:=cost[k,j];
closest[j]:=k;
end;
end;
end;{prim}

function find(v:integer):integer; {返回頂點(diǎn)v所在的集合}
var i:integer;
begin
i:=1;
while (i<=n) and (not v in vset[i]) do inc(i);
if i<=n then find:=i else find:=0;
end; 

procedure kruskal;
var
tot,i,j:integer;
begin
for i:=1 to n do vset[i]:=[i];{初始化定義n個(gè)集合，第I個(gè)集合包含一個(gè)元素I}
p:=n-1; q:=1; tot:=0; {p為尚待加入的邊數(shù)，q為邊集指針}
sort;
{對(duì)所有邊按權(quán)值遞增排序，存于e[I]中，e[I].v1與e[I].v2為邊I所連接的兩個(gè)頂點(diǎn)的序號(hào)，e[I].len為第I條邊的長度}
while p>0 do begin
i:=find(e[q].v1);j:=find(e[q].v2);
if i<>j then begin
inc(tot,e[q].len);
vset[i]:=vset[i]+vset[j];vset[j]:=[];
dec(p);
end;
inc(q);
end;
writeln(tot);
end;

var
a:array[1..maxn,1..maxn] of integer;
b:array[1..maxn] of integer; {b[i]指頂點(diǎn)i到源點(diǎn)的最短路徑}
mark:array[1..maxn] of boolean; 

procedure bhf;
var
best,best_j:integer;
begin
fillchar(mark,sizeof(mark),false);
mark[1]:=true; b[1]:=0;{1為源點(diǎn)}
repeat
best:=0;
for i:=1 to n do
If mark[i] then {對(duì)每一個(gè)已計(jì)算出最短路徑的點(diǎn)}
for j:=1 to n do
if (not mark[j]) and (a[i,j]>0) then 
if (best=0) or (b[i]+a[i,j]<best) then begin
best:=b[i]+a[i,j]; best_j:=j;
end;
if best>0 then begin
b[best_j]:=best；mark[best_j]:=true;
end;
until best=0;
end;{bhf}

procedure floyed;
begin
for I:=1 to n do
for j:=1 to n do
if a[I,j]>0 then p[I,j]:=I else p[I,j]:=0; {p[I,j]表示I到j(luò)的最短路徑上j的前驅(qū)結(jié)點(diǎn)}
for k:=1 to n do {枚舉中間結(jié)點(diǎn)}
for i:=1 to n do
for j:=1 to n do
if a[i,k]+a[j,k]<a[i,j] then begin
a[i,j]:=a[i,k]+a[k,j];
p[I,j]:=p[k,j];
end;
end; 

var
a:array[1..maxn,1..maxn] of integer;
b,pre:array[1..maxn] of integer; {pre[i]指最短路徑上I的前驅(qū)結(jié)點(diǎn)}
mark:array[1..maxn] of boolean;
procedure dijkstra(v0:integer);
begin
fillchar(mark,sizeof(mark),false);
for i:=1 to n do begin
d[i]:=a[v0,i];
if d[i]<>0 then pre[i]:=v0 else pre[i]:=0;
end;
mark[v0]:=true;
repeat {每循環(huán)一次加入一個(gè)離1集合最近的結(jié)點(diǎn)并調(diào)整其他結(jié)點(diǎn)的參數(shù)}
min:=maxint; u:=0; {u記錄離1集合最近的結(jié)點(diǎn)}
for i:=1 to n do
if (not mark[i]) and (d[i]<min) then begin
u:=i; min:=d[i];
end;
if u<>0 then begin
mark[u]:=true; 
for i:=1 to n do
if (not mark[i]) and (a[u,i]+d[u]<d[i]) then begin
d[i]:=a[u,i]+d[u];
pre[i]:=u;
end;
end;
until u=0;
end;

Procedure Longlink;
Var
T:array[1..maxn,1..maxn] of boolean;
Begin
Fillchar(t,sizeof(t),false);
For k:=1 to n do
For I:=1 to n do
For j:=1 to n do T[I,j]:=t[I,j] or (t[I,k] and t[k,j]);
End; 

procedure dfs ( now,color: integer);
begin
for i:=1 to n do
if a[now,i] and c[i]=0 then begin {對(duì)結(jié)點(diǎn)I染色}
c[i]:=color;
dfs(I,color);
end;
end; 

x[I],y[I],t[I]分別表示第I條邊的起點(diǎn)，終點(diǎn)和權(quán)。共n個(gè)結(jié)點(diǎn)和m條邊。
procedure bellman-ford
begin
for I:=0 to n-1 do d[I]:=+infinitive;
d[0]:=0;
for I:=1 to n-1 do
for j:=1 to m do {枚舉每一條邊}
if d[x[j]]+t[j]<d[y[j]] then d[y[j]]:=d[x[j]]+t[j];
for I:=1 to m do
if d[x[j]]+t[j]<d[y[j]] then return false else return true;
end; 

procedure search(k,v:integer); {搜索第k個(gè)物品，剩余空間為v}
var i,j:integer;
begin
if v<best then best:=v;
if v-(s[n]-s[k-1])>=best then exit; {s[n]為前n個(gè)物品的重量和}
if k<=n then begin
if v>w[k] then search(k+1,v-w[k]);
search(k+1,v);
end;
end; 

l DP
F[I,j]為前i個(gè)物品中選擇若干個(gè)放入使其體積正好為j的標(biāo)志，為布爾型。
實(shí)現(xiàn):將最優(yōu)化問題轉(zhuǎn)化為判定性問題
f [I, j] = f [ i-1, j-w[i] ] (w[I]<=j<=v) 邊界：f[0,0]:=true.
For I:=1 to n do
For j:=w[I] to v do F[I,j]:=f[I-1,j-w[I]];
優(yōu)化：當(dāng)前狀態(tài)只與前一階段狀態(tài)有關(guān)，可降至一維。
F[0]:=true;
For I:=1 to n do begin
F1:=f;
For j:=w[I] to v do
If f[j-w[I]] then f1[j]:=true;
F:=f1;
End;

Procedure update;
var j,k:integer;
begin
c:=a;
for j:=0 to n do
if a[j]>0 then
if j+now<=n then inc(c[j+now],a[j]);
a:=c;
end; 

Begin
FillChar(f,SizeOf(f),0);
For i:=1 To M Do
For j:=1 To N Do
If i-problem[j].time>=0 Then
Begin
t:=problem[j].point+f[i-problem[j].time];
If t>f[i] Then f[i]:=t;
End;
Writeln(f[M]);
End. 

procedure try(dep:integer);
var i,j:integer;
begin
cal; {此過程計(jì)算當(dāng)前系數(shù)的計(jì)算結(jié)果，now為結(jié)果}
if now>n then exit; {剪枝}
if dep=l+1 then begin {生成所有系數(shù)}
cal;
if now=n then inc(tot);
exit;
end;
for i:=0 to n div pr[dep] do begin
xs[dep]:=i;
try(dep+1);
xs[dep]:=0;
end;
end; 

procedure try(dep,rest:integer);
var i,j,x:integer;
begin
if (rest<=0) or (dep=l+1) then begin
if rest=0 then inc(tot);
exit;
end;
for i:=0 to rest div pr[dep] do
try(dep+1,rest-pr[dep]*i);
end;
{main: try(1,n); } 

Procedure update;
var j,k:integer;
begin
c:=a;
for j:=0 to n do
if a[j]>0 then
for k:=1 to n div now do
if j+now*k<=n then inc(c[j+now*k],a[j]);
a:=c;
end;
{main}
begin 
read(now); {讀入第一個(gè)物品的重量}
i:=0; {a[i]為背包容量為i時(shí)的放法總數(shù)}
while i<=n do begin 
a[i]:=1; inc(i,now); end; {定義第一個(gè)物品重的整數(shù)倍的重量a值為1，作為初值}
for i:=2 to v do
begin
read(now);
update; {動(dòng)態(tài)更新}
end;
writeln(a[n]);

procedure qsort(l,r:integer);
var i,j,mid:integer;
begin
i:=l;j:=r; mid:=a[(l+r) div 2]; {將當(dāng)前序列在中間位置的數(shù)定義為中間數(shù)}
repeat
while a[i]<mid do inc(i); {在左半部分尋找比中間數(shù)大的數(shù)}
while a[j]>mid do dec(j);{在右半部分尋找比中間數(shù)小的數(shù)}
if i<=j then begin {若找到一組與排序目標(biāo)不一致的數(shù)對(duì)則交換它們}
swap(a[i],a[j]);
inc(i);dec(j); {繼續(xù)找}
end;
until i>j;
if l<j then qsort(l,j); {若未到兩個(gè)數(shù)的邊界，則遞歸搜索左右區(qū)間}
if i<r then qsort(i,r);
end;{sort} 

procedure insert_sort;
var i,j:integer;
begin
for i:=2 to n do begin
a[0]:=a[i];
j:=i-1;
while a[0]<a[j] do begin
a[j+1]:=a[j];
j:=j-1;
end;
a[j+1]:=a[0];
end;
end;{inset_sort} 

procedure sort;
var i,j,k:integer;
begin
for i:=1 to n-1 do 
for j:=i+1 to n do
if a[i]>a[j] then swap(a[i],a[j]);
end; 

procedure bubble_sort;
var i,j,k:integer;
begin
for i:=1 to n-1 do
for j:=n downto i+1 do
if a[j]<a[j-1] then swap( a[j],a[j-1]); {每次比較相鄰元素的關(guān)系}
end; 

procedure sift(i,m:integer);{調(diào)整以i為根的子樹成為堆,m為結(jié)點(diǎn)總數(shù)}
var k:integer;
begin
a[0]:=a[i]; k:=2*i;{在完全二叉樹中結(jié)點(diǎn)i的左孩子為2*i,右孩子為2*i+1}
while k<=m do begin
if (k<m) and (a[k]<a[k+1]) then inc(k);{找出a[k]與a[k+1]中較大值}
if a[0]<a[k] then begin a[i]:=a[k];i:=k;k:=2*i; end
else k:=m+1;
end;
a[i]:=a[0]; {將根放在合適的位置}
end; 

procedure heapsort;
var
j:integer;
begin
for j:=n div 2 downto 1 do sift(j,n);
for j:=n downto 2 do begin
swap(a[1],a[j]);
sift(1,j-1);
end;
end; 

var I,j,t:integer;
tmp:listtype;
begin
t:=p;i:=p;j:=q+1;{t為tmp指針，I,j分別為左右子序列的指針}
while (t<=r) do begin
if (i<=q){左序列有剩余} and ((j>r) or (a[i]<=a[j])) {滿足取左邊序列當(dāng)前元素的要求}
then begin
tmp[t]:=a[i]; inc(i);
end
else begin
tmp[t]:=a[j];inc(j);
end;
inc(t);
end;
for i:=p to r do a[i]:=tmp[i];
end;{merge} 

procedure merge_sort(var a:listtype; p,r: integer); {合并排序a[p..r]}
var q:integer;
begin
if p<>r then begin
q:=(p+r-1) div 2;
merge_sort (a,p,q);
merge_sort (a,q+1,r);
merge (a,p,q,r);
end;
end;
{main}
begin
merge_sort(a,1,n);
end. 

procedure plus ( a,b:hp; var c:hp);
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
if a[0]>b[0] then len:=a[0] else len:=b[0];
for i:=1 to len do begin
inc(c[i],a[i]+b[i]);
if c[i]>10 then begin dec(c[i],10); inc(c[i+1]); end; {進(jìn)位}
end;
if c[len+1]>0 then inc(len);
c[0]:=len;
end;{plus} 

procedure substract(a,b:hp;var c:hp); 
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
if a[0]>b[0] then len:=a[0] else len:=b[0];
for i:=1 to len do begin
inc(c[i],a[i]-b[i]);
if c[i]<0 then begin inc(c[i],10);dec(c[i+1]); end;
while (len>1) and (c[len]=0) do dec(len);
c[0]:=len;
end; 

procedure multiply(a:hp;b:longint;var c:hp);
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
len:=a[0];
for i:=1 to len do begin
inc(c[i],a[i]*b);
inc(c[i+1],(a[i]*b) div 10);
c[i]:=c[i] mod 10;
end;
inc(len);
while (c[len]>=10) do begin {處理最高位的進(jìn)位}
c[len+1]:=c[len] div 10;
c[len]:=c[len] mod 10;
inc(len);
end;
while (len>1) and (c[len]=0) do dec(len); {若不需進(jìn)位則調(diào)整len}
c[0]:=len;
end;{multiply} 

procedure high_multiply(a,b:hp; var c:hp}
var i,j,len:integer;
begin
fillchar(c,sizeof(c),0);
for i:=1 to a[0] do
for j:=1 to b[0] do begin
inc(c[i+j-1],a[i]*b[j]);
inc(c[i+j],c[i+j-1] div 10);
c[i+j-1]:=c[i+j-1] mod 10;
end;
len:=a[0]+b[0]+1;
while (len>1) and (c[len]=0) do dec(len);
c[0]:=len;
end; 

procedure devide(a:hp;b:longint; var c:hp; var d:longint);
{c:=a div b; d:= a mod b}
var i,len:integer;
begin
fillchar(c,sizeof(c),0);
len:=a[0]; d:=0;
for i:=len downto 1 do begin
d:=d*10+a[i];
c[i]:=d div b;
d:=d mod b;
end;
while (len>1) and (c[len]=0) then dec(len);
c[0]:=len;
end; 

procedure high_devide(a,b:hp; var c,d:hp);
var
i,len:integer;
begin
fillchar(c,sizeof(c),0);
fillchar(d,sizeof(d),0);
len:=a[0];d[0]:=1;
for i:=len downto 1 do begin
multiply(d,10,d);
d[1]:=a[i];
while(compare(d,b)>=0) do {即d>=b}
begin
Subtract(d,b,d);
inc(c[i]);
end;
end;
while(len>1)and(c.s[len]=0) do dec(len);
c.len:=len;
end; 

procedure Solve(pre,mid:string);
var i:integer;
begin
if (pre='''') or (mid='''') then exit;
i:=pos(pre[1],mid);
solve(copy(pre,2,i),copy(mid,1,i-1));
solve(copy(pre,i+1,length(pre)-i),copy(mid,i+1,length(mid)-i));
post:=post+pre[1]; {加上根，遞歸結(jié)束后post即為后序遍歷}
end; 

procedure Solve(mid,post:string);
var i:integer;
begin
if (mid='''') or (post='''') then exit;
i:=pos(post[length(post)],mid);
pre:=pre+post[length(post)]; {加上根，遞歸結(jié)束后pre即為前序遍歷}
solve(copy(mid,1,I-1),copy(post,1,I-1));
solve(copy(mid,I+1,length(mid)-I),copy(post,I,length(post)-i));
end; 

function ok(s1,s2:string):boolean;
var i,l:integer; p:boolean;
begin
ok:=true;
l:=length(s1);
for i:=1 to l do begin
p:=false;
for j:=1 to l do
if s1[i]=s2[j] then p:=true;
if not p then begin ok:=false;exit;end;
end;
end; 

procedure solve(pre,post:string);
var i:integer;
begin
if (pre='''') or (post='''') then exit;
i:=0;
repeat
inc(i);
until ok(copy(pre,2,i),copy(post,1,i));
solve(copy(pre,2,i),copy(post,1,i));
midstr:=midstr+pre[1];
solve(copy(pre,i+2,length(pre)-i-1),copy(post,i+1,length(post)-i-1));
end; 

procedure solve(dep:integer);
var
i:integer;
begin
if dep=n+1 then begin writeln(s);exit; end;
for i:=1 to n do
if not used[i] then begin
s:=s+chr(i+ord(''0''));used[i]:=true;
solve(dep+1);
s:=copy(s,1,length(s)-1); used[i]:=false;
end;
end; 

procedure solve(dep,pre:integer);
var
i:integer;
begin
if dep=k+1 then begin writeln(s);exit; end;
for i:=1 to n do
if (not used[i]) and (i>pre) then begin
s:=s+chr(i+ord(''0''));used[i]:=true;
solve(dep+1,i);
s:=copy(s,1,length(s)-1); used[i]:=false;
end;
end; 

function binsearch(k:keytype):integer;
var low,hig,mid:integer;
begin
low:=1;hig:=n;
mid:=(low+hig) div 2;
while (a[mid].key<>k) and (low<=hig) do begin
if a[mid].key>k then hig:=mid-1
else low:=mid+1;
mid:=(low+hig) div 2;
end;
if low>hig then mid:=0;
binsearch:=mid;
end; 

function treesrh(k:keytype):pointer;
var q:pointer;
begin
q:=root;
while (q<>nil) and (q^.key<>k) do
if k<q^.key then q:=q^.left
else q:=q^.right;
treesrh:=q;
end; 

procedure try(i:byte);
var j:byte;
begin
if i=n+1 then begin print;exit;end;
for j:=1 to n do
if a[i] and b[j+i] and c[j-i] then begin
x[i]:=j;
a[j]:=false; b[j+i]:=false; c[j-i]:=false;
try(i+1);
a[j]:=true; b[i+j]:=true; c[j-i]:=true;
end;
end;

h(n)=2*h(n-1)+1 
h(1)=1
初始所有銅片都在a柱上
procedure hanoi(n,a,b,c:byte); {將第n塊銅片從a柱通過b柱移到c柱上}
begin
if n=0 then exit;
hanoi(n-1,a,c,b); {將上面的n-1塊從a柱通過c柱移到b柱上}
write(n,'moved from',a,'to',c);
hanoi(n-1,b,a,c);{ 將b上的n-1塊從b柱通過a柱移到c柱上
end; 

Procedure move(k,goal:integer); {將最大不到位的k移到目標(biāo)柱goal上}
Begin
If k=0 then exit;
For I:=1 to 3 do
For j:=1 to han[I,0] do
If h[I,j]=k then begin now:=I;nowp:=j; end; {找到k的位置}
If now<>goal then begin {若未移到目標(biāo)}
Move(k-1,6-now-goal); {剩下的先移到?jīng)]用的柱上}
Writeln(k moved from now to goal);
H[goal,h[goal,0]+1]:=h[now,nowp]; h[now,nowp]:=0;
Inc(h[goal,0]); dec(h[now,0]);
Move(k-1,goal); {剩下的移到目標(biāo)上}
End;

procedure work(dep,pre,s:longint); {入口為work(1,1,n)}
{dep為當(dāng)前試放的第dep個(gè)數(shù),pre為前一次試放的數(shù),s為當(dāng)前剩余可分的總數(shù)}
var j:longint;
begin 
if dep=n then begin 
if s>=pre then inc(r); exit; 
end; 
for j:=pre to s div 2 do work(dep+1,j,s-j); 
end;
類似：
procedure try(dep:integer);
var i:integer;
begin
if dep=k then begin
if tot>=a[dep-1] then inc(sum);
exit; end;
for i:=a[dep-1] to tot div 2 do begin
a[dep]:=i; dec(tot,i); 
try(dep+1);
inc(tot,i);
end;
end;{try} 

head:=1; tail:=0;
while tail<head do begin
inc(tail);
for k:=1 to n do
if k方向可擴(kuò)展 then begin
inc(head);
list[head].x:=list[tail].x+dx[k]; {擴(kuò)展出新結(jié)點(diǎn)list[head]}
list[head].y:=list[tail].y+dy[k];
處理新結(jié)點(diǎn)list[head];
end;
end; 

loc(I:integer):pointer; {尋找鏈表中的第I個(gè)結(jié)點(diǎn)的指針}
procedure loc(L:linklist; I:integer):pointer;
var p:pointer;
j:integer;
begin
p:=L.head; j:=0;
if (I>=1) and (I<=L.len) then 
while j<I do begin p:=p^.next; inc(j); end;
loc:=p;
end; 

procedure insert(L:linklist; I:integer; x:datatype);
var p,q:pointer;
begin
p:=loc(L,I);
new(q);
q^.data:=x;
q^.next:=p^.next;
p^.next:=q;
inc(L.len);
end; 

procedure delete(L:linklist; I:integer);
var p,q:pointer;
begin
p:=loc(L,I-1);
q:=p^.next;
p^.next:=q^.next;
dispose(q);
dec(L.len);
end; 

p:=loc(L,I);
new(q);
q^.data:=x;
q^.pre:=p;
q^.next:=p^.next;
p^.next:=q;
q^.next^.pre:=q; 

p:=loc(L,I); {p為要?jiǎng)h除的結(jié)點(diǎn)}
p^.pre^.next:=p^.next;
p^.next^.pre:=p^.pre;
dispose(p);