jz2638 【提高】移动棋子

 

 

分析：
   有规律可寻。

 

const maxn=1000;
var
    b:array[1..maxn]of char;
    sp,i,n,num:longint;

procedure print;
var i:longint;
begin
    inc(num);
    write('step ',num,':');
    for i:=1 to 2*(n+1) do write(b[i]); 
    writeln;
end;

procedure move(k:longint);
var i:longint;
begin
    for i:=0 to 1 do  begin b[sp+i]:=b[k+i];b[k+i]:='_'; end;
    sp:=k; 
    print;
end;

procedure mv(n:longint);
begin
    if n=4 then
    begin 
        move(4); move(8); move(2); move(7); move(1);
    end
    else
    begin 
        move(n); move(2*n-1); mv(n-1);
    end;
end;

begin
    readln(n);
    for i:=1 to n do b[i]:='O';
    for i:=n+1 to 2*n do b[i]:='*';
    b[i+1]:='_'; b[i+2]:='_';
    sp:=2*n+1;
    num:=-1;
    print;
    mv(n);
end.

 

 

#include<cstring>
#include<cstdio>
#include<iostream>
using namespace std;

string s,ans;
int n;
int main()
{
    cin>>n;
    s="";
    for (int i=0;i<n;i++) s+='O';
    for (int i=0;i<n;i++) s+='*';
    s+="__";
    int x=n,tot=0;
    cout<<"step "<<tot<<":"<<s<<endl;
    tot++;
    while (x>4)
    {
        s[x-1]='_'; s[x]='_'; s[2*x]='O'; s[2*x+1]='*';
        cout<<"step "<<tot<<":"<<s<<endl;
        tot++;
        s[2*x-1]='_'; s[2*x-2]='_'; s[x-1]='*'; s[x]='*';
        cout<<"step "<<tot<<":"<<s<<endl;
        x--;
        tot++;
    }
    ans=s.substr(10,(n-4)*2);
    cout<<"step "<<tot<<":"<<"OOO__***O*"+ans<<endl; tot++;
    cout<<"step "<<tot<<":"<<"OOO*O**__*"+ans<<endl; tot++;
    cout<<"step "<<tot<<":"<<"O__*O**OO*"+ans<<endl; tot++;
    cout<<"step "<<tot<<":"<<"O*O*O*__O*"+ans<<endl; tot++;
    cout<<"step "<<tot<<":"<<"__O*O*O*O*"+ans<<endl; tot++;
    return 0;
}

 

var s,ans:string;
      n,total,i,x:longint;
begin
  readln(n);
  for i:=1 to n do
    s:=s + 'O';
  for i:=1 to n do
    s:=s + '*';
  s:=s + '__';
  x:=n;
  writeln('step ',total,':',s);
  inc(total);
  while x > 4 do
  begin
      s[x]:='_'; s[x+1]:='_'; s[2*x+1]:='O'; s[2*x+2]:='*';
      writeln('step ',total,':',s);
      inc(total);
      s[2*x]:='_'; s[2*x-1]:='_'; s[x]:='*'; s[x+1]:='*';
      dec(x);
      writeln('step ',total,':',s);
      inc(total);
  end;
  ans:=copy(s,11,(n-4)*2);
  writeln('step ',total,':OOO__***O*'+ans); inc(total);
  writeln('step ',total,':OOO*O**__*'+ans); inc(total);
  writeln('step ',total,':O__*O**OO*'+ans); inc(total);
  writeln('step ',total,':O*O*O*__O*'+ans); inc(total);
  writeln('step ',total,':__O*O*O*O*'+ans); inc(total);
end.

 

转载于:https://www.cnblogs.com/qilinart/articles/4963291.html