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2011-2012 集训队第一次集中学习

2011-2012 集训队第一次集中学习. 进制转换. repeat i: ＝ i ＋１； ｙ mod ｘ 的余数──→ a[i] ） ｙ := ｙ div ｘ Until ｙ＝０. 要考虑的问题，如果是10以上的进制，余数有可能是2位数. program ex; var n,nx:real; nz:integer; procedure changez(nz:integer); var x:integer; begin x:=nz mod 2; nz:=nz div 2;

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2011-2012 集训队第一次集中学习

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  1. 2011-2012集训队第一次集中学习 进制转换

  2. repeat i:＝i＋１； ｙ mod ｘ 的余数──→ a[i]） ｙ:= ｙ div ｘ Until ｙ＝０ 要考虑的问题，如果是10以上的进制，余数有可能是2位数

  3. program ex; var n,nx:real; nz:integer; procedure changez(nz:integer); var x:integer; begin x:=nz mod 2; nz:=nz div 2; if nz<>0 then changez(nz); write(x); end; procedure changex(nx:real); var i,t:integer; begin for i:=1 to 5 do begin nx:=nx*2; t:=trunc(nx); write(t); nx:=nx-t; end; end; begin readln(n); nz:=trunc(n); nx:=n-nz; changez(nz); write('.'); changex(nx); end.

  4. procedure changex(t:integer); var i,j,q,x:integer; begin for i:=1 to 5 do begin q:=0; for j:=1 to len2 do begin x:=b[j]*2+q; q:=x div 10; b[j]:=x mod 10; end; write(q); end; end; program zh_10_2; const w=5; var st,st1,st2:string; a,b,c:array[0..1000] of integer; p,len,len1,len2,i:integer; procedure changez(t:integer); var i,j,x:integer; begin j:=0; while t<>0 do begin inc(j); x:=0; for i:= t downto 1 do begin a[i]:=x*10+a[i]; x:=a[i] mod 2; a[i]:=a[i] div 2; end; c[j]:=x; while (a[t]=0) and (t>0) do dec(t); end; for i:=j downto 1 do write(c[i]); end; 4455666565565.65656565 begin readln(st); len:=length(st); p:=pos('.',st); st1:=copy(st,1,p-1); st2:=copy(st,p+1,len-p); len1:=length(st1); len2:=length(st2); for i:=1 to 1000 do begin a[i]:=0 ;b[i] :=0;c[i]:=0; end; for i:=1 to len1 do a[i]:=ord(st1[len1-i+1])-48; for i:=1 to len2 do b[i]:=ord(st2[len2-i+1])-48; changez(len1); write('.'); changex(len2); end.

  5. procedure changez(t:integer); var i,j,x:integer; begin j:=0; while t<>0 do begin inc(j); x:=0; for i:= t downto 1 do begin a[i]:=x*10+a[i]; x:=a[i] mod 2; a[i]:=a[i] div 2; end; c[j]:=x; while (a[t]=0) and (t>0) do dec(t); end; for i:=j downto 1 do write(c[i]); end; 2 1 5 3 1 8 6 7 0 7 6 5 9 8 3 1

  6. procedure changex(t:integer); var i,j,q,x:integer; begin for i:=1 to 5 do begin q:=0; for j:=1 to len2 do begin x:=b[j]*2+q; q:=x div 10; b[j]:=x mod 10; end; write(q); end; end; 6 5 6 5 6 5 6 5 2 ×

  7. 进制转换K进制 function dectok(x,k:longint):string; const alph='0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'; var st:string; begin st:=''; while x<>0 do begin st:=alph[x mod k+1]+st; x:=x div k; end; exit(st); end;

  8. K进制转换十进制 function ktodec(st:string; k:longint):longint; const alph='012456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'; var i,j,ans:longint; begin ans:=0; j:=1; for i:=length(st) downto 1 do begin inc(ans,j*(pos(st[i],alph)-1)); j:=j*k; end; exit(ans); end;

  9. 题1128：基础二进制小数 题1258：【基础】任意进制转换

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