2133 - [数值问题]高精度乘以低精度
【问题描述】
输入两个高精度正整数a和b(a位数<=200,b的位数<=9位),求两数的乘积。
【输入格式】lowmul.in
<span>输入共两行,分别为a和b。</span>
<span>【输出格式】lowmul.out</span>
<span>输出共一行,表示两个数的积。</span>
<span>【输入样例1】 </span><span><br />
9999999999999999999999999999999999
<span style="line-height:21px;">1111111</span>
<span>【输出样例1】</span>
99999999899999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999000000001
题目输入
题目输出
输入/输出样例
输入格式
输出格式
提示
Const
SIZE = 200+10;
Type
hugeint = Record
len : Integer;
num : Array[1..SIZE] Of int64;
End;
var a:hugeint;
s1:string;
i:integer;
x:longint;
procedure lowmul(a:hugeint;x:longint);
var
i:longint;
ans:hugeint;
begin
fillchar(ans.num,sizeof(ans.num),0);
ans.len:=a.len;
for i:=1 to ans.len do ans.num[i]:=a.num[i]*x;
for i:=1 to ans.len do
begin
inc(ans.num[i+1],ans.num[i] div 10);
ans.num[i]:=ans.num[i] mod 10;
end;
while ans.num[ans.len+1]>0 do
begin
inc(ans.len);
inc(ans.num[ans.len+1],ans.num[ans.len] div 10);
ans.num[ans.len]:=ans.num[ans.len] mod 10;
end;
for i:=ans.len downto 1 do write(ans.num[i]);
writeln;
end;
procedure datain;
begin
assign(input,'lowmul.in'); assign(output,'lowmul.out');
reset(input); rewrite(output);
readln(s1);
readln(x);
a.len:=length(s1);
for i:=1 to a.len do a.num[i]:=ord(s1[a.len-i+1])-ord('0');
end;
begin
datain;
lowmul(a,x);
close(input);close(output);
end.
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