NOIP模板

快排

procedure qsort(l,r:longint);
var i,j,t,m:longint;
begin i:=l; j:=r;
 m:=a[(i+j) div 2];
 repeat
while a[i]<m do inc(i);
while m<a[j] do dec(j);
if i<=j then
  begin   t:=a[i]; a[i]:=a[j]; a[j]:=t;
inc(i); dec(j);  end;  until i>j;
 if i<r then qsort(i,r); if j>l then qsort(l,j);  end;  

 

1.数组开两倍

procedure insert(x:longint);
var t,s,v:longint;
begin
  len:=len+1; f[len]:=x; s:=len;
  while (s<>1)and(f[s div 2]>f[s]) do
   begin
     v:=f[s div 2]; f[s div 2]:=f[s]; f[s]:=v;
     s:=s div 2;
   end;
end;
function get:longint;
var t,s,v:longint;
begin
  get:=f[1]; f[1]:=f[len]; len:=len-1; t:=1;
  while (t*2<=len)or(t*2+1<=len) do
   begin
     if (t*2+1>len)or(f[t*2]<f[t*2+1])
     then s:=t*2 else  s:=t*2+1;
     if f[t]>f[s] then
      begin
        v:=f[t]; f[t]:=f[s]; f[s]:=v; t:=s;
      end
  else break;
   end;
end;    

 

 

欧拉函数(单个数值)

function phi(x:longint):longint;
var ans,i:longint;
begin
  ans:=x;
  for i:=2 to trunc(sqrt(x)) do
   begin
   if x mod i=0 then ans:=ans div i*(i-1);
   while x mod i=0 do x:=x div i;
   end;
  if x<>1 then ans:=ans div x*(x-1);
  exit(ans);
end;        

欧拉函数(批量处理)

procedure work;
var i,j:longint;
begin
  p:=0;
  for i:=0 to n do e[i]:=i;
  for i:=2 to n do
    if e[i]=i then
     begin j:=i;
     while j<=n do begin e[j]:=e[j] div i*(i-1); inc(j,i); end;
     end;
  for i:=1 to n do
   begin
     if e[i]=i-1 then begin inc(p); prime[p]:=i; end;
   end;
end;

质数筛法

procedure work(n:longint);
var i,j:longint;
begin
  num:=0;
  for i:=1 to n do g[i]:=true;
  for i:=2 to n do
   if g[i]=true then
    begin
      inc(num); prime[num]:=i;
      j:=i*2;
      while j<=n do begin g[j]:=false; j:=j+i; end;
    end;
end;  

质因数分解

procedure fenjie(x:longint);
var i,y:longint;
begin
  t:=0;y:=x;
  for i:=1 to num do
   begin
     if prime[i]>y div 2 then exit;
     if x mod prime[i]=0 then
     begin
     inc(t); w[t]:=prime[i];
     while x mod prime[i]=0 do x:=x div prime[i];
     end;
     if x=1 then break;
   end;
end;   

 

快速幂

function quick(x,n,p:qword):qword;
var ans:qword;
begin
  ans:=1;
  while n>0 do
  begin
   if n and 1>0 then ans:=cheng(ans,x,p);
   x:=cheng(x,x,p);
   n:=n>>1;
  end;
  exit(ans);
end;

快速乘

function cheng(x,n,p:qword):qword;
var ans:qword;
begin
 ans:=0;
  while n>0 do
   begin
     if n and 1>0 then ans:=(ans+x) mod p;
     x:=(x+x) mod p;
     n:=n>>1;
   end;
  exit(ans);
end;

 

拓扑排序

for i:=1 to n do if u[i]=0 then begin inc(t); f[t]:=i; end;
  repeat
    x:=f[t]; dec(t); inc(num);
    new(p); p:=a[x];
    while p<>nil do
     begin
       y:=p^.data;  
       dec(u[y]);
       if u[y]=0 then
        begin
          inc(t); f[t]:=y;
        end;
       p:=p^.next;
     end;
  until num=n;

 

 

强连通分量tarjan

思路:dfn为遍历到各点的时间,low表示每个点能够追溯到的最早的栈中节点的次序号 instack表示结点是否在栈中。

对图遍历如果x的儿子y没访问过,则访问并用low[y]更行low[x],如果x的儿子y被访问过,则用dfn[y]更新low[x],这样就知道了x最早回溯到的点(在栈中),在栈中low[x]相等的值构成强连通分量。

procedure tarjan(x:longint);
var y:longint; p:point;
begin
  inc(s); low[x]:=s; dfn[x]:=s; inc(t); q[t]:=x;
  instack[x]:=true;
  new(p); p:=a[x];
  while p<>nil do
   begin
     y:=p^.x;
     if dfn[y]=0 then begin tarjan(y); low[x]:=min(low[x],low[y]); end
      else if instack[y]=true then low[x]:=min(low[x],dfn[y]);
     p:=p^.next;
   end;
  if dfn[x]=low[x] then
   begin
     inc(num);
     repeat
       y:=q[t]; dec(t); f[y]:=x;instack[y]:=false;
     until x=y;
   end;
end;

主程序

 for i:=1 to n do begin dfn[i]:=0; low[i]:=0; w[i]:=0; instack[i]:=false; end;
  s:=0; t:=0; num:=0;
  for i:=1 to n do if dfn[i]=0 then tarjan(i);     

 

 

 

树的直径

思路:求树上最长路,对于各个点,求出它儿子中最长和次长的链相加。

procedure dfs(x,e:longint);
var p:point; y:longint;
begin
  s1[x]:=x; s2[x]:=x; new(p); p:=a[x];
  while p<>nil do
   begin
     y:=p^.x;
     if y=e then begin p:=p^.next; continue; end;
     dfs(y,x);
     if f1[y]+p^.v>f1[x] then
     begin
       f2[x]:=f1[x]; s2[x]:=s1[x];
       f1[x]:=f1[y]+p^.v; s1[x]:=y;
      end
     else
     if f1[y]+p^.v>f2[x] then
     begin
       f2[x]:=f1[y]+p^.v; s2[x]:=y;
     end;
     p:=p^.next;
   end;
  if f1[x]+f2[x]>f1[max]+f2[max] then
   max:=x;
end;       

 

kmp算法

  p[1]:=0; j:=0;
  for i:=2 to m do
   begin
     while (j>0)and(b[j+1]<>b[i]) do j:=p[j];
     if b[j+1]=b[i] then inc(j);
     p[i]:=j;
   end;
  j:=0; kmp:=0;
  for i:=1 to n do
    begin
      while (j>0)and(b[j+1]<>a[i]) do j:=p[j];
      if b[j+1]=a[i] then inc(j);
      if j=m then begin inc(kmp); j:=p[j]; end;
    end;

 

 

LCA

倍增

1.不要忘记dfs后调用work

2.注意枚举2^i时要包含0,注意顺序

procedure work;
var i,j:longint;
begin
  for i:=0 to 19 do
   for j:=1 to n do
    f[i+1,j]:=f[i,f[i,j]];
end;
function lca(x,y:longint):longint;
var i,t:longint;
begin
  if deep[x]<deep[y] then begin t:=x; x:=y; y:=t; end;
  for i:=19 downto 0 do
   if ((deep[x]-deep[y])>>i)and 1=1 then x:=f[i,x];
  if x=y then exit(x);
  for i:=19 downto 0 do
   if f[i,x]<>f[i,y] then
   begin x:=f[i,x]; y:=f[i,y]; end;
  exit(f[0,x]);
end;    

 

二分图

染色

function dfs(x,color:longint):boolean;
var p:point; y:longint;
begin
  new(p); p:=w[x]; g[x]:=color;
  while p<>nil do
   begin
     y:=p^.x;
     if g[y]=color then exit(false);
     if (g[y]=0)and(not dfs(y,-color)) then exit(false);
     p:=p^.next;
   end;
  exit(true);
end;
function cheak(s:longint):boolean;
var i:longint;
begin
  for i:=1 to n do g[i]:=0;
  for i:=1 to n do
   if g[i]=0 then
    if not dfs(i,1) then exit(false);
  exit(true);
end;

 组合

费马小定理,快速求C(n,m)的值

procedure make;
var i:longint;
begin
  f[0]:=1; g[0]:=1;
  for i:=1 to n do
   begin
     f[i]:=(f[i-1]*i) mod num;
     g[i]:=g[i-1]*quick(i,num-2) mod num;
   end;
end;
function get(x,y:int64):int64;
var i:longint; t:int64;
begin
  if (x=0)or(y=0)or(x=y) then exit(1);
  exit(((f[x]*g[x-y]) mod num)*g[y] mod num);
end;

 

同余

ax≡1(mod m)

program asdf;
var
  a,m,x,y,t:longint;
function extgcd(a,b:longint;var x,y:longint):longint;
var d:longint;
begin
  if b<>0 then begin d:=extgcd(b,a mod b,y,x); y:=y-(a div b)*x; end
   else begin x:=1;y:=0; d:=a; end;
  exit(d);
end;
begin
  readln(a,m);
  x:=0; y:=0;
  t:=extgcd(a,m,x,y);
  writeln((x+m) mod m);
end.
posted @ 2016-09-16 16:18  QTY_YTQ  阅读(300)  评论(0编辑  收藏  举报