Pascal master enters, will pascal enter

The scale of this data is very small, and the simulation is enough, and it can be bubbling if it is sorted.

var n:integer; m,ans:real;

a:array[1..100]of integer;

procedure swap(x,y:integer);

var k;

begink:=x;

x:=y;y:=k;

end;begin

read(n);

for i:=1 to n do read(a[i]);

for i:=1 to n-1 do

for j:=i to n-1 do

if a[j]m:=0;for i:=2 to n-1 do inc(m,a[i]);

ans:=m/(n-2);

write(ans);

end.I can't remember the operation function of the specific shaping transfer, you can search it on the Internet, and you can use snapshots or stacking to improve efficiency.

Of course there is: word dword qword

Respectively, the leakage is an unsigned integer longint int64

The range is less than twice the front of the signed variable, and the silver search is 0 65535 for word for example

Problem analysis: Know the range of stamps per letter (<=3), there are four types of stamps, and program to find the stamps that can make the largest face value. The algorithm is:

1) For four stamps of different denominations, no more than 3 stamps shall be affixed to each letter;

2) Use these four stamps to affix consecutive integers and maximize the r value;

3) Use exhaustive methods to find all eligible solutions;

4) This question uses the method of collection to count the face value of stamps to improve the speed of weighting.

The face values of the four stamps are: A, B, C, D, according to the title

ax0 then

beginx0:=x;x1:=a;x2:=b;x3:=c;x4:=d;

write(x1:5,x2:5,x3:5,x4:5);

writeln(‘’10,’x0=’,x0);

end;end;

end.

Do your own online search for information about BFS.

It's a very simple question, why are you taking it out?

Any questions?

In the future, if there is any problem with the program, come to me.

You can right-click and select Send desktop shortcut.

Then copy the shortcut to the desktop, click Start, All Programs, Launch, and then copy the shortcut into it.

This question is actually a slight variation of the Joseph problem, program ysf;

typearr1=record

next:integer;

data:boolean;

name:string;

end;var

a:array[1..64] of arr1;

n,m,i,j,k,p,w:integer;

temp:string;

beginreadln(n);

for i:=1 to n-1 do

with a[i] do

begin next:=i+1;data:=true;readln(name);end;

with a[n] do

begin next:=1;data:=true;readln(name);end;

readln(temp);

val(copy(temp,1,pos(',temp)-1),w,p);

val(copy(temp,pos(',temp)+1,length(temp)-pos(',temp)),m,p);

i:=w;k:=1;p:=n;

repeat

repeat

i:=a[i].next;

until a[i].data;

inc(k);

if k=m then

begin a[i].data:=false;dec(p);writeln(a[i].name,' k:=0;end;

until p=0;

end.

Posterior traversal: left, right, center.

ch stores the number of each read, a[i] is the i power of 2 (a[n] is the total length of the chain of seepage), t represents the character length of the current tree, how many factors 2 contains in i represents the current node is the penultimate layer (each layer node is the t number from i-t+1 i, the current number is the rightmost node of this layer), s iterates each number in the current node, if the sum is t, then it proves that all are 1, so it is i tree; If it's 0, it's a b number, otherwise it's an f-tree.

If the human brain simulates it, you can find that this algorithm is completely in the order of left, right, and middle, so the traversal is followed by the post-order traversal.

Do you know how to operate the tree?

Flowers that don't recursively use a [0..10,1..2] can also be used for array simulation.