高精度计算25925.doc

高精度计算 2.1 高精度加法2.2 高精度减法2.3 高精度乘法 2.4 高精度除法 2.1高精度加法 高精度加法程序如下： program HighPrecision1_Plus;const? fn_inp=hp1.inp;? fn_out=hp1.out;? maxlen=100;? { max length of the number }type? hp=record?????? len:integer; { length of the number }?????? s:array[1..maxlen] of integer?????? { s[1]?? is the lowest? position???????? s[len] is the highest position }???? end;var? x:array[1..2] of hp;? y:hp; { x:input ; y:output } ? procedure PrintHP(const p:hp);? var i:integer;? begin??? for i:=p.len downto 1 do write(p.s[i]);? end; ? procedure init;? var??? st:string;??? j,i:integer;? begin??? assign(input,fn_inp);??? reset(input);??? for j:=1 to 2 do??? begin????? readln(st);????? x[j].len:=length(st);????? for i:=1 to x[j].len do { change string to HP }??????? x[j].s[i]:=ord(st[x[j].len+1-i])-ord(0);??? end;??? close(input);? end; ? procedure Plus(a,b:hp;var c:hp); { c:=a+b }? var i,len:integer;? begin??? fillchar(c,sizeof(c),0);??? if a.lenb.len then len:=a.len? { get the bigger length of a,b }?????????????????? else len:=b.len;??? for i:=1 to len do { plus from low to high }??? begin????? inc(c.s[i],a.s[i]+b.s[i]);????? if c.s[i]=10 then????? begin??????? dec(c.s[i],10);??????? inc(c.s[i+1]); { add 1 to a higher position }????? end;??? end;??? if c.s[len+1]0 then inc(len);??? c.len:=len;? end; ? procedure main;? begin??? Plus(x[1],x[2],y);? end; ? procedure out;? begin??? assign(output,fn_out);??? rewrite(output);??? PrintHP(y);??? writeln;??? close(output);? end; ? begin??? init;??? main;??? out;? end. 2. 2 高精度减法 高精度减法程序如下： program HighPrecision2_Subtract;const? fn_inp=hp2.inp;? fn_out=hp2.out;? maxlen=100;? { max length of the number }type? hp=record?????? len:integer; { length of the number }?????? s:array[1..maxlen] of integer?????? { s[1]?? is the lowest? position???????? s[len] is the highest position }???? end;var? x:array[1..2] of hp;? y:hp; { x:input ; y:output }? positive:boolean