石破天惊 @ 2016-01-28 22:56:04
const met=200;
var
i,j,la,lb,len:longint;
sa,sb:string;
m:longint;
a:array[1..met]of longint;
b:array[1..met]of longint;
c:array[1..met]of longint;
begin
for i:=1 to met do c[i]:=0;
readln(sa);
readln(sb);
la:=length(sa);
for i:=1 to la do a[i]:=ord(sa[la-i+1])-48;
lb:=length(sb);
for i:=1 to lb do b[i]:=ord(sb[lb-i+1])-48;
for i:=1 to la do
for j:=1 to lb do
c[i+j-1]:=c[i+j-1]+a[i]*b[j];
len:=la+lb;
for i:=1 to len do
begin
c[i+1]:=c[i+1]+c[i] div 10;
c[i]:=c[i] mod 10;
end;
while c[len]=0 do dec(len);
m:=c[len];
while m>0 do
begin
c[len]:=m mod 10;
m:=m div 10;
inc(len);
end;
for i:=len-1 downto 1 do
write(c[i]);
readln;
end.by 石破天惊 @ 2016-01-28 23:13:12
求助/i
by 石破天惊 @ 2016-01-28 23:14:26
[codep ]writeln("Hello world!")[/codep ]
by 汪力宾 @ 2016-01-30 16:14:31
var
s1,s2:ansistring;
a,b,c:array[1..100000] of integer;
jw,la,lb,i,j,ws:longint;
begin
readln(s1);
readln(s2);
if (s1='0')or(s2='0') then
begin
writeln(0);
halt;
end;
la:=length(s1);
lb:=length(s2);
for i:=1 to la do
a[i]:=ord(s1[la-i+1])-48;
for i:=1 to lb do
b[i]:=ord(s2[lb-i+1])-48;
for i:=1 to lb do
begin
jw:=0;
for j:=1 to la+1 do
begin
c[i+j-1]:=c[i+j-1]+a[j]*b[i]+jw;
jw:=c[i+j-1] div 10;
c[i+j-1]:=c[i+j-1] mod 10;
end;
end;
ws:=la+lb;
if c[ws]=0 then
dec(ws);
for i:=ws downto 1 do
write(c[i]);
end.
by 汪力宾 @ 2016-01-30 16:15:13
自己看吧
by xunzhen @ 2016-02-02 21:09:36
水题一道，我用的C语言(压位)。
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
char s[30000],l[30000];
long long a[30001],b[30001],c[60000],al,bl,cl,i,j,x;
int main()
{gets(s);
al=strlen(s);
for (i=0;i<(5-al%5)%5;i++){
strcpy(l,s);
strcpy(s,"0");
strcat(s,l);
}
//puts(s);
if (al%5!=0) al=(al/5+1)*5;
for (i=0;i<al/5;i++)
a[i+1]=(s[al-i*5-1]-'0')+(s[al-i*5-2]-'0')*10+(s[al-i*5-3]-'0')*100+(s[al-i*5-4]-'0')*1000+(s[al-i*5-5]-'0')*10000;
gets(s);
bl=strlen(s);
for (i=0;i<(5-bl%5)%5;i++){
strcpy(l,s);
strcpy(s,"0");
strcat(s,l);
}
if (bl%5!=0) bl=(bl/5+1)*5;
for (i=0;i<bl/5;i++)
b[i+1]=(s[bl-i*5-1]-'0')+(s[bl-i*5-2]-'0')*10+(s[bl-i*5-3]-'0')*100+(s[bl-i*5-4]-'0')*1000+(s[bl-i*5-5]-'0')*10000;
for (i=1;i<=bl/5;i++)
for (j=1;j<=al/5;j++)
c[i+j-1]=c[i+j-1]+a[j]*b[i];
for (i=1;i<=al/5+bl/5;i++){
c[i+1]=c[i+1]+c[i]/100000;
c[i]=c[i]%100000;
}
x=0;
for (i=al/5+bl/5;i>0;i--){
if (c[i]!=0 && x==0){
x=1; cl=i;
}
if (x){
if(i<cl){
if (c[i]<10) printf("0");
if (c[i]<100) printf("0");
if (c[i]<1000) printf("0");
if (c[i]<10000) printf("0");
}
printf("%d",c[i]);
}
}
if (x==0) printf("0");
puts("");
system("pause");
return 0;
}
by 饕餮吃题目 @ 2016-08-10 18:53:50
我直接写sum：=a*b都有四十分哈哈哈！
by YYX6 @ 2016-08-14 15:38:48
@caiyuelin我也是
by wangzh @ 2016-08-16 08:53:14
#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
int x[10001]={0},y[10001]={0},z[10001]={0};//一定要加上={0}
int main()
{
int temp;
string a,b;
int l=0;
cin>>a>>b;
int c=a.size()-1,d=b.size()-1;
for(int i=0;i<=c;i++)
x[i]=a[i]-48;
for(int i=0;i<=d;i++)
y[i]=b[i]-48;
for(int i=0;i<=c/2;i++)
{
temp=x[i];
x[i]=x[c-i];
x[c-i]=temp;
}
for(int i=0;i<=d/2;i++)
{
temp=y[i];
y[i]=y[d-i];
y[d-i]=temp;
}
for(int i=0;i<=c;i++)
for(int j=0;j<=d;j++)
{
z[i+j]=x[i]*y[j]+z[i+j];
if(z[i+j]>9)
{
z[j+i+1]+=z[j+i]/10;
z[j+i]-=z[j+i]/10*10;
}
l=i+j;
}
l++;
while(z[l]==0&&l!=0)l--;
for(int i=l;i>=0;i--)
cout<<z[i];
return 0;
}by wangzh @ 2016-08-16 08:53:55
自己看吧，C++的