高精度计算练习2

Posted 2020-06-13

高精度运算_乘法

AYYZOJ p1445

 1 program p1445;
 2 const
 3   max=5000;
 4 var
 5  a,b,c:array[1..max] of word;
 6  n1,n2:string;
 7  lena,lenb,lenc,i,j,x:integer;
 8 begin
 9  readln(n1);
10  readln(n2);
11  lena:=length(n1); lenb:=length(n2);
12  for i:=1 to lena do a[lena-i+1]:=ord(n1[i])-ord(‘0‘);
13  for i:=1 to lenb do b[lenb-i+1]:=ord(n2[i])-ord(‘0‘);
14  for i:=1 to lena do
15  begin
16   x:=0;
17   for j:=1 to lenb do
18   begin
19    C[i+j-1]:= A[i]*B[j]+ x + C[i+j-1];
20    x:= C[i+j-1] div 10;
21    C[i+j-1]:= C[i+j-1] mod 10
22   end;
23   c[i+j]:= x;
24  end;
25  lenc:=i+j;
26  while (c[lenc]=0) and (lenc>1) do dec(lenc);
27  for i:=lenc downto 1 do write(c[i]);
28 end.

高精度乘法

高精度运算_A/B问题1

AYYZOJ p1446

 1 program p1446;
 2 const e=20;
 3 var
 4  a,d,x:array[0..e] of integer;
 5  n,m,t:integer;
 6 begin
 7  read(n,m);
 8  write(n,‘/‘,m,‘=‘);
 9  a[0]:=n; d[0]:=n div m;
10  x[0]:=n mod m;
11  write(d[0],‘.‘);
12   for t:=1 to e do
13   begin
14    //if x[t-1]=0 then exit; {去掉也能A}
15    a[t]:=x[t-1]*10;
16    d[t]:=a[t] div m;
17    write(d[t]);
18    x[t]:=a[t] mod m;
19   end;
20 end.

View Code

高精度运算_A/B问题2

AYYZOJ p1447

 1 const max=200;dot=100;
 2 var a,c:array[0..max] of byte;
 3    x,n2:longint; n1:string;
 4    lena:integer; code,i,j:integer;
 5 begin
 6   readln(n1);
 7   readln(n2);
 8   lena:=length(n1);
 9   for i:=1 to lena do a[i] := ord(n1[i]) - ord(‘0‘);
10   x:=0;
11   for i:=1 to lena do begin
12     c[i]:=(x*10+a[i]) div n2;
13     x:=(x*10+a[i]) mod n2;
14   end;
15   j:=1;
16   while (c[j]=0) and (j<lena) do inc(j);
17   if j<>1 then
18   for i:=j to lena do begin c[i-j+1]:=c[i];c[i]:=0;end;
19   c[0]:=lena-j+1;
20   if x=0 then begin
21     for i:=1 to c[0] do write(c[i]);
22     writeln;halt;
23   end;
24   j:=c[0]+dot;
25   for i:=1 to dot do begin
26     c[c[0]+i]:=x*10 div n2;
27     x:=x*10 mod n2;
28     if x=0 then begin j:=c[0]+i;break;end;
29   end;
30   for i:=1 to j do
31     if i=c[0] then write(c[i],‘.‘)
32      else write(c[i]);
33   writeln
34 end.

View Code

高精度运算_3的n次幂
AYYZOJ p1450

 1 program p1450;
 2 var
 3   a:array[1..500] of byte;
 4   i,j,c,n,l:integer;
 5 begin
 6   readln(n);
 7   a[1]:=1;l:=1;
 8   for i:=1 to n do
 9     begin
10       c:=0;
11       for j:=1 to l do
12         begin
13           a[j]:=a[j]*3+c;
14           c:=a[j] div 10;
15           a[j]:=a[j] mod 10;
16         end;
17       if c>0 then begin inc(l);a[l]:=c;end;
18     end;
19   for i:=l downto 1 do write(a[i]);
20   writeln;
21 end.

View Code

高精度运算_求s=1+2+...+n

AYYZOJ p1451

 1 var
 2   n:longint;
 3   a,b,c:array[1..100] of integer;
 4   la,lb,lc,k,i,j:integer;
 5 procedure mul(x,y:longint);
 6   begin
 7     la:=0;
 8     while x>0 do
 9       begin
10         inc(la);
11         a[la]:=x mod 10;
12         x:=x div 10;
13       end;
14     lb:=0;
15     while y>0 do
16       begin
17         inc(lb);
18         b[lb]:=y mod 10;
19         y:=y div 10;
20       end;
21     for i:=1 to la do
22       begin
23         k:=0;
24         for j:=1 to lb do
25           begin
26             c[i+j-1]:=a[i]*b[j]+k+c[i+j-1];
27             k:=c[i+j-1] div 10;
28             c[i+j-1]:=c[i+j-1] mod 10;
29           end;
30         c[i+j]:=k;
31       end;
32     lc:=i+j;
33     while(c[lc]=0) and (lc>1) do dec(lc);
34     for i:=lc downto 1 do write(c[i]);writeln;
35   end;
36 begin
37   readln(n);
38   if n mod 2=0 then mul(n div 2,n+1)  {等差数列求和公式}
39   else mul(n,(n+1) div 2);
40 end.

View Code

高精度运算_求n的阶乘

AYYZOJ p1452

 1 const maxn=3001;maxm=10000;
 2 type arr=array[1..maxm] of integer;
 3 var
 4   ans:arr;
 5   p,num:array[1..maxn] of integer;
 6   n,i,j,k:longint;
 7   flag:boolean;la:integer;
 8 procedure add(x:longint);
 9   var i:longint;
10   begin
11     i:=1;
12     while p[i]<=x do begin
13       while (x mod p[i]=0) do
14         begin
15           inc(num[i],1);
16           x:=x div p[i];
17         end;
18       inc(i);
19     end;
20   end;
21 procedure mul(var a:arr;c:longint;var la:integer);
22   var i:longint;
23   begin
24     a[1]:=a[1]*c;
25     for i:=2 to la do
26       begin
27         a[i]:=a[i]*c;
28         a[i]:=a[i]+a[i-1] div 10;
29         a[i-1]:=a[i-1] mod 10;
30       end;
31     while a[la]>=10 do
32       begin
33         inc(la);
34         a[la]:=a[la-1] div 10;
35         a[la-1]:=a[la-1] mod 10;
36       end;
37   end;
38 begin
39   p[1]:=2;k:=1;
40   for i:=3 to maxn do begin
41     flag:=false;
42     for j:=2 to trunc(sqrt(i)) do
43       if i mod j=0 then begin flag:=true;break;end;
44     if not flag then begin inc(k);p[k]:=i;end;
45   end;
46   fillchar(ans,sizeof(ans),0);
47   fillchar(num,sizeof(num),0);
48   readln(n);
49   for i:=2 to n do add(i);
50   ans[1]:=1; la:=1;
51   for i:=1 to maxn do
52     if p[i]>n then break else for j:=1 to num[i] do mul(ans,p[i],la);
53   for i:=la downto 1 do write(ans[i]);writeln;
54 end.

优化举例

N!

 1 const m=10000000;
 2 var a:array[1..m] of integer;
 3     f:real;
 4     y,c,i,j,n,k,l,t:longint;
 5     st1,st2:string;
 6 begin
 7   write(‘input a number n:‘);
 8   read(n);
 9  // for i:=1 to n do
10  // f:=f+ln(i)/ln(10);
11  // t:=trunc(f)+1;            {计算可能需要的位数}
12   a[1]:=1;l:=1;k:=1;
13   for i:=1 to n do
14    begin
15     c:=0;
16     while a[k]=0 do k:=k+1;
17     for j:=k to l do
18      begin
19       y:=a[j]*i+c;            {该位上的数乘以I加上低位的进位}
20       c:=y div 10000;            {每四位数占用1个数组单元,c是进位}
21       a[j]:=y mod 10000
22      end;
23     if c>0 then begin
24               l:=l+1;
25               a[l]:=c;            {若有进位则数组的下标指针加1,增加一}
26              end;                    {个存储单元}
27   end;
28   write(n,‘!=‘);
29   for i:=l downto 1 do            {输出处理}
30    begin
31      if i=l then write(a[i])
32             else begin
33                   str(a[i],st1);
34                   while length(st1)<4 do
35                   st1:=‘0‘+st1;
36                   write(st1);
37                  end;
38    end;
39   writeln;
40 end.

N! pascal（优化举例）