vj p1024题解

原题叙述

此题很简单，搜索题。

按照步骤老老实实的做就能过，不过这题要注意要用int64，高精也可以。虽然题目要求的是说maxlongint以内，但是由于倒过来就不只maxlongint了。

另：记得补0和自己本身有可能是循环节。

代码如下：

 

var  n,k,r:int64;
    g,i,j,l:longint;
    a: array [ 0 .. 10000 ] of  int64;
    over:boolean;
    t:string;
    num: array [ 0 .. 9 ] of  longint;
procedure  make(g:longint;n:int64; var  a,b:int64);
           var  s:string;i,j:longint;z:int64;
begin
     fillchar(num,sizeof(num), 0 );
     str(n,s);
      for  i: = 1   to  length(s)  do
          begin
              inc(num[ord(s[i]) - ord( ' 0 ' )]);
          end ;
     z: = 1 ;
     a: = 0 ;
     i: = 0 ;
      for  i: = 0   to   9   do
          begin
               if  num[i] <> 0   then
                  begin
                       for  j: = 1   to  num[i]  do
                           begin
                               a: = a + i * z;
                               z: = z * 10 ;
                           end ;
                  end ;
          end ;
     z: = 1 ;
     b: = 0 ;
      for  i: = 9   downto   0   do
          begin
               if  num[i] <> 0   then
                  begin
                       for  j: = 1   to  num[i]  do
                           begin
                               dec(num[i]);
                               b: = b + i * z;
                               z: = z * 10 ;
                           end ;
                  end ;
          end ;
end ;
begin
      while   not (eof)  do
            begin
                readln(n);
                str(n,t);
                g: = length(t);
                make(g,n,k,r);
                r: = k - r;
                a[ 1 ]: = n;
                l: = 2 ;
                a[ 2 ]: = r;
                over: = false;
                 while   not (over)  do
                       begin
                           make(g,a[l],k,r);
                           inc(l);
                           a[l]: = k - r;
                            for  i: = l - 1   downto   1   do
                                begin
                                     if  a[l] = a[i]  then
                                        begin
                                             for  j: = i  to  l - 1   do  write(a[j], '   ' );
                                            writeln;
                                            over: = true;
                                            break;
                                        end ;
                                end ;
                       end ;
            end ;
end .

 

转载于:https://www.cnblogs.com/waterfalleagle/archive/2009/11/06/1597721.html