进制1

【题目】键盘输入一个十进制的整数，及确定进制n，把这个数转换成相应的n进制输出。

（其中２〈＝n〈＝１６）

例如：输入１０，n＝３ 则输出 (10)10=(101)3

【参考程序】

var i,j,m,n:longint;

a:array [1..100] of byte; {用数组存放求出的余数}

begin

writeln('input m,n');

fillchar(a,sizeof(a),0);

readln(m,n);

write(m);

i:=0;

repeat

i:=i+1;

a[i]:= m mod n;

m:=m div n;

until m=0;

write('=(');

for j:=i downto 1 do

if a[j]>9 then write(chr(a[j]+55)) {如果大于９，用字母输出}

else write(a[j]);

writeln(')',n);readln;

end.

.context 进制2

【题目】把n进制的数化回十进制表示

如 (10101)2=(21)10

【参考程序】

var cf,s,i,j,n:longint;

m:string[20];

a:array[1..20] of byte;

begin

writeln('input m,n');

fillchar(a,sizeof(a),0);

readln(m,n); {用字符串接收要转换的数}

for i:=1 to length(m) do begin {把字符串换成数字,注意字母时的情况}

if (m[i]<='9') and (m[i]>='0') then a[i]:=ord(m[i])-48;

if (upcase(m[i])<='F') and (upcase(m[i])>='A') then

a[i]:=ord(upcase(m[i]))-55;

if a[i]>=n then begin writeln('Error, Invaild m !');halt;end;

{如果含有不在n进制内的字符，则判为出错。如２进制的数，则不应

出现诸如10102,110031210等情况}

end;

cf:=1; s:=a[length(m)]; {cf:乘方}

for i:=length(m)-1 downto 1 do begin {从低位向高位，逐步转换}

cf:=cf*n; {s记录得出来的数}

s:=s+a[i]*cf;

end;

writeln('(',m,')',n,'=',s);

readln;

end.

.context 进制3

【题目】任意进制间的互化。

把n进制的Ｍ转化成k进制表示

如m=ff n=16 k=2

则有 (ff)16=(11111111)2

【参考程序】

var s,n,k:longint;m:string[20];

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

procedure first(m:string;n:integer); {把数m化成十进制}

var cf,i,j:longint;

begin

for i:=1 to length(m) do begin

if (m[i]<='9') and (m[i]>='0') then a[i]:=ord(m[i])-48;

if (upcase(m[i])<='F') and (upcase(m[i])>='A') then

a[i]:=ord(upcase(m[i]))-55;

if a[i]>=n then begin writeln('Error, Invaild m !');halt;end;

end;

cf:=1; s:=a[length(m)];

for i:=length(m)-1 downto 1 do begin

cf:=cf*n;

s:=s+a[i]*cf;

end;

write('(',m,')',n);write('=(',s,')10');

end;

procedure second(m,n:longint); {把十进制的数化成k进制}

var i,j:longint;

begin

i:=0;

repeat

i:=i+1;

a[i]:= m mod n;

m:=m div n;

until m=0;

write('=(');

for j:=i downto 1 do

if a[j]>9 then write(chr(a[j]+55))

else write(a[j]);

writeln(')',n);readln;

end;

begin

fillchar(a,sizeof(a),0);

writeln('input m,n,k:'); {m:数, n:原先进制, k:化成什么进制}

readln(m,n,k);

first(m,n); {把n进制的m化成十进制数Ｓ}

second(s,k); {把Ｓ化成k进制}

end.

进制2

【题目】把n进制的数化回十进制表示

如 (10101)2=(21)10

【参考程序】

var cf,s,i,j,n:longint;

m:string[20];

a:array[1..20] of byte;

begin

writeln('input m,n');

fillchar(a,sizeof(a),0);

readln(m,n); {用字符串接收要转换的数}

for i:=1 to length(m) do begin {把字符串换成数字,注意字母时的情况}

if (m[i]<='9') and (m[i]>='0') then a[i]:=ord(m[i])-48;

if (upcase(m[i])<='F') and (upcase(m[i])>='A') then

a[i]:=ord(upcase(m[i]))-55;

if a[i]>=n then begin writeln('Error, Invaild m !');halt;end;

{如果含有不在n进制内的字符，则判为出错。如２进制的数，则不应

出现诸如10102,110031210等情况}

end;

cf:=1; s:=a[length(m)]; {cf:乘方}

for i:=length(m)-1 downto 1 do begin {从低位向高位，逐步转换}

cf:=cf*n; {s记录得出来的数}

s:=s+a[i]*cf;

end;

writeln('(',m,')',n,'=',s);

readln;

end.

.context 进制3

【题目】任意进制间的互化。

把n进制的Ｍ转化成k进制表示

如m=ff n=16 k=2

则有 (ff)16=(11111111)2

【参考程序】

var s,n,k:longint;m:string[20];

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

procedure first(m:string;n:integer); {把数m化成十进制}

var cf,i,j:longint;

begin

for i:=1 to length(m) do begin

if (m[i]<='9') and (m[i]>='0') then a[i]:=ord(m[i])-48;

if (upcase(m[i])<='F') and (upcase(m[i])>='A') then

a[i]:=ord(upcase(m[i]))-55;

if a[i]>=n then begin writeln('Error, Invaild m !');halt;end;

end;

cf:=1; s:=a[length(m)];

for i:=length(m)-1 downto 1 do begin

cf:=cf*n;

s:=s+a[i]*cf;

end;

write('(',m,')',n);write('=(',s,')10');

end;

procedure second(m,n:longint); {把十进制的数化成k进制}

var i,j:longint;

begin

i:=0;

repeat

i:=i+1;

a[i]:= m mod n;

m:=m div n;

until m=0;

write('=(');

for j:=i downto 1 do

if a[j]>9 then write(chr(a[j]+55))

else write(a[j]);

writeln(')',n);readln;

end;

begin

fillchar(a,sizeof(a),0);

writeln('input m,n,k:'); {m:数, n:原先进制, k:化成什么进制}

readln(m,n,k);

first(m,n); {把n进制的m化成十进制数Ｓ}

second(s,k); {把Ｓ化成k进制}

end.

进制3

【题目】任意进制间的互化。

把n进制的Ｍ转化成k进制表示

如m=ff n=16 k=2

则有 (ff)16=(11111111)2

【参考程序】

var s,n,k:longint;m:string[20];

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

procedure first(m:string;n:integer); {把数m化成十进制}

var cf,i,j:longint;

begin

for i:=1 to length(m) do begin

if (m[i]<='9') and (m[i]>='0') then a[i]:=ord(m[i])-48;

if (upcase(m[i])<='F') and (upcase(m[i])>='A') then

a[i]:=ord(upcase(m[i]))-55;

if a[i]>=n then begin writeln('Error, Invaild m !');halt;end;

end;

cf:=1; s:=a[length(m)];

for i:=length(m)-1 downto 1 do begin

cf:=cf*n;

s:=s+a[i]*cf;

end;

write('(',m,')',n);write('=(',s,')10');

end;

procedure second(m,n:longint); {把十进制的数化成k进制}

var i,j:longint;

begin

i:=0;

repeat

i:=i+1;

a[i]:= m mod n;

m:=m div n;

until m=0;

write('=(');

for j:=i downto 1 do

if a[j]>9 then write(chr(a[j]+55))

else write(a[j]);

writeln(')',n);readln;

end;

begin

fillchar(a,sizeof(a),0);

writeln('input m,n,k:'); {m:数, n:原先进制, k:化成什么进制}

readln(m,n,k);

first(m,n); {把n进制的m化成十进制数Ｓ}

second(s,k); {把Ｓ化成k进制}

end.