1754: [Usaco2005 qua]Bull Math
1754: [Usaco2005 qua]Bull Math
HansBug
发布于 2018-04-11 10:31:12
发布于 2018-04-11 10:31:12
1754: [Usaco2005 qua]Bull Math
Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 398 Solved: 242
Description
Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... or so they say. Farmer John wonders if their answers are correct. Help him check the bulls' answers. Read in two positive integers (no more than 40 digits each) and compute their product. Output it as a normal number (with no extra leading zeros). FJ asks that you do this yourself; don't use a special library function for the multiplication. 输入两个数,输出其乘积
Input
* Lines 1..2: Each line contains a single decimal number.
Output
* Line 1: The exact product of the two input lines
Sample Input
11111111111111 1111111111
Sample Output
12345679011110987654321
HINT
Source
题解：萌萌哒高精度乘法= =
1 /**************************************************************
2 Problem: 1754
3 User: HansBug
4 Language: Pascal
5 Result: Accepted
6 Time:0 ms
7 Memory:376 kb
8 ****************************************************************/
9
10 var
11 i,j,k,l,m,n:longint;
12 a,b,c:array[0..10000] of longint;
13 s1,s2:ansistring;
14 begin
15 readln(s1);
16 readln(s2);
17 a[0]:=length(s1);
18 for i:=1 to a[0] do a[i]:=ord(s1[a[0]+1-i])-48;
19 b[0]:=length(s2);
20 for i:=1 to b[0] do b[i]:=ord(s2[b[0]+1-i])-48;
21 c[0]:=a[0]+b[0];
22 for i:=1 to a[0] do
23 for j:=1 to b[0] do
24 begin
25 c[i+j-1]:=c[i+j-1]+a[i]*b[j];
26 c[i+j]:=c[i+j]+c[i+j-1] div 10;
27 c[i+j-1]:=c[i+j-1] mod 10;
28 end;
29 while (c[0]>1) and (c[c[0]]=0) do dec(c[0]);
30 for i:=c[0] downto 1 do write(c[i]);
31 writeln;
32 readln;
33 end. 本文参与 腾讯云自媒体分享计划，分享自作者个人站点/博客。
原始发表：2015-04-14 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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