6426: [数值问题]高精度乘以高精度
[命题人 : ]
题目描述
【问题描述】
输入两个高精度正整数a和b(a,b位数<=200),求两数的乘积。
【输入格式】highmul.in 输入共两行,分别为a和b。
【输出格式】highmul.out
输出共一行,表示两个数的积。
【输入样例1】
1234567890
1234567890
【输出样例1】
1524157875019052100
样例输入 复制
样例输出 复制
提示
Const
SIZE = 400;
Type
hugeint = Record
len : Integer;
num : Array[1..SIZE] Of integer;
End;
var a,b:hugeint;
s1,s2:string;
i:integer;
procedure times(a, b : hugeint);
Var
i, j : Integer;
ans : hugeint;
Begin
FillChar(ans, SizeOf(ans), 0);
For i := 1 To a.len Do
For j := 1 To b.len Do
ans.num[i + j - 1] := ans.num[i + j - 1] + a.num[i] * b.num[j];
For i := 1 To a.len + b.len Do
Begin
ans.num[i + 1] := ans.num[i + 1] + ans.num[i] DIV 10;
ans.num[i] := ans.num[i] mod 10;
If ans.num[a.len + b.len] > 0
Then ans.len := a.len + b.len
Else ans.len := a.len + b.len - 1;
End;
for i:=ans.len downto 1 do write(ans.num[i]);
writeln;
End;
procedure datain;
begin
assign(input,'highmul.in'); assign(output,'highmul.out');
reset(input); rewrite(output);
readln(s1);
readln(s2);
a.len:=length(s1);
b.len:=length(s2);
for i:=1 to a.len do a.num[i]:=ord(s1[a.len-i+1])-ord('0');
for i:=1 to b.len do b.num[i]:=ord(s2[b.len-i+1])-ord('0');
end;
begin
datain;
times(a,b);
close(input);
close(output);
end.
SIZE = 400;
Type
hugeint = Record
len : Integer;
num : Array[1..SIZE] Of integer;
End;
var a,b:hugeint;
s1,s2:string;
i:integer;
procedure times(a, b : hugeint);
Var
i, j : Integer;
ans : hugeint;
Begin
FillChar(ans, SizeOf(ans), 0);
For i := 1 To a.len Do
For j := 1 To b.len Do
ans.num[i + j - 1] := ans.num[i + j - 1] + a.num[i] * b.num[j];
For i := 1 To a.len + b.len Do
Begin
ans.num[i + 1] := ans.num[i + 1] + ans.num[i] DIV 10;
ans.num[i] := ans.num[i] mod 10;
If ans.num[a.len + b.len] > 0
Then ans.len := a.len + b.len
Else ans.len := a.len + b.len - 1;
End;
for i:=ans.len downto 1 do write(ans.num[i]);
writeln;
End;
procedure datain;
begin
assign(input,'highmul.in'); assign(output,'highmul.out');
reset(input); rewrite(output);
readln(s1);
readln(s2);
a.len:=length(s1);
b.len:=length(s2);
for i:=1 to a.len do a.num[i]:=ord(s1[a.len-i+1])-ord('0');
for i:=1 to b.len do b.num[i]:=ord(s2[b.len-i+1])-ord('0');
end;
begin
datain;
times(a,b);
close(input);
close(output);
end.