其实，除了Python,Haskell也可以一行

main = do { a<-readLn; b<-readLn; print (a*b) }

golang大法
```go

packagepa main

import (
"fmt"
"math/big"
)

func main() {

var stra, strb string

fmt.Scanf("%s\n%s", &stra, &strb)

a := big.NewInt(0)
b := big.NewInt(0)
ans := big.NewInt(0)

a.SetString(stra, 10)
b.SetString(strb, 10)

ans.Mul(a, b)

fmt.Println(ans)
}

#include<bits/stdc++.h>
using namespace std;
int main(){
    char st1[10005],st2[10005];
    int a[10005],b[10005],c[20005];
    int i,j,len1,len2;
    cin >> st1 >> st2;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    len1=strlen(st1);
    len2=strlen(st2);
    j=1;
    for (i=len1-1;i>=0;i--)
    {
        a[j++]=st1[i]-'0';
    }
    j=1;
    for (i=len2-1;i>=0;i--)
    {
        b[j++]=st2[i]-'0';
    }

    for (i=1;i<=len1;i++)
        for (j=1;j<=len2;j++)
            c[i+j-1]+=a[i]*b[j];

    for(i=1;i<len1+len2;i++){
        c[i+1]+=c[i]/10;
        c[i]%=10;
    }
    while (!c[i] && i>1){
        i--;
    }
    while (i){
        printf("%d",c[i--]);
    }
    printf("\n");
    return 0;
}

在网上淘得代码，确实是很不错，模拟你现实用笔算的过程，在计算机中模拟出来，真心给个赞。

#include<bits/stdc++.h>
using namespace std;
int main()
{
char a1[30000],a2[30000];
int a[30000],b[30000],c[30000],la,lb,lc,i,j,x;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
cin>>a1>>a2;
la=strlen(a1);
lb=strlen(a2);
for(i=0;i<=la-1;i++) a[la-i]=a1[i]-48;
for(i=0;i<=lb-1;i++) b[lb-i]=a2[i]-48;
for(i=1;i<=la;i++)
{
x=0;
for(j=1;j<=lb;j++)
{
c[i+j-1]=a[i]*b[j]+x+c[i+j-1];
x=c[i+j-1]/10;
c[i+j-1]%=10;
}
c[i+lb]=x;
}
lc=la+lb;
while(c[lc]==0&&lc>1)
lc--;
for(i=lc;i>=1;i--)
cout<<c[i];
cout<<endl;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
char a1[30000],a2[30000];
int a[30000],b[30000],c[30000],la,lb,lc,i,j,x;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
cin>>a1>>a2;
la=strlen(a1);
lb=strlen(a2);
for(i=0;i<=la-1;i++) a[la-i]=a1[i]-48;
for(i=0;i<=lb-1;i++) b[lb-i]=a2[i]-48;
for(i=1;i<=la;i++)
{
x=0;
for(j=1;j<=lb;j++)
{
c[i+j-1]=a[i]*b[j]+x+c[i+j-1];
x=c[i+j-1]/10;
c[i+j-1]%=10;
}
c[i+lb]=x;
}
lc=la+lb;
while(c[lc]==0&&lc>1)
lc--;
for(i=lc;i>=1;i--)
cout<<c[i];
cout<<endl;
}

Java大法好

import java.math.BigInteger;
import java.util.Scanner;

public class Main 
{
    public static void main(String[] args) 
    {
        Scanner in = new Scanner(System.in);
        BigInteger A = in.nextBigInteger(), B = in.nextBigInteger();
        System.out.println(A.multiply(B));
        in.close();
    }
}

print(int(input())int(input()))

print(int(input())*int(input()))

//没有压位。。。蜜汁过了lol
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <string>

using namespace std;

const int MAXN = 10005;
string s1, s2;

struct BigInt{
    int num[MAXN * 4], size;
    BigInt(string str){
        size = (int)str.size();
        for (int i = 0; i < str.size(); i++){num[i] = str[size - i - 1] - '0';}
    }
    BigInt operator*(BigInt b2){
        const int *num2 = b2.num;
        BigInt neu("");
        neu.size = b2.size + size;
        for (int i = 0; i < this -> size; i++){
            for (int j = 0; j < b2.size; j++){
                neu.num[i + j] += num2[j] * num[i];
            }
        }
        for (int i = 1; i < neu.size + 10; i++){
            neu.num[i] += neu.num[i - 1] / 10;
            neu.num[i - 1] %= 10;
        }
        for (int i = neu.size + 10; i >= 0; i--){
            if (neu.num[i] != 0) {neu.size = i + 1; break;}
        }
        return neu;
    }
    void print(){
        for (int i = this -> size - 1; i >= 0; i--){
            printf("%d", this -> num[i]);
        }
        printf("\n");
    }
};

int main(){
    cin >> s1 >> s2;
    BigInt b1(s1), b2(s2);
    BigInt neu = b1 * b2;
    neu.print();
    return 0;
}

print(int(input()) * int(input()))

人生苦短(

FFT

#include<iostream>
#include<cmath>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<cstring>
using namespace std;
const long double pi=acos(-1);
struct cp{long double x,y;};
cp operator +(cp a,cp b){return (cp){a.x+b.x,a.y+b.y};}
cp operator -(cp a,cp b){return (cp){a.x-b.x,a.y-b.y};}
cp operator *(cp a,cp b){return (cp){a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};}
cp cur[1000005],a[1000005],b[1000005];
char ch[1000005];
int len1,len2,len,ans[1000005];
void fft(cp *a,int n,int fl){
for(int i=(n>>1),j=1;j<n;++j){
if(i<j)swap(a[i],a[j]);
int k;
for(k=(n>>1);k&i;i^=k,k>>=1);i^=k;
}
for(int m=2;m<=n;m<<=1){
cp w=(cp){cos(2*pi*fl/m),sin(2*pi*fl/m)};
cur[0]=(cp){1,0};
for(int j=1;j<(m>>1);j++)cur[j]=cur[j-1]*w;
for(int i=0;i<n;i+=m)
for(int j=i;j<i+(m>>1);++j){
cp u=a[j],v=a[j+(m>>1)]*cur[j-i];
a[j]=u+v;
a[j+(m>>1)]=u-v;
}
}
if(fl==-1)
for(int i=0;i<n;i++)
a[i]=(cp){a[i].x/n,a[i].y/n};
}
int main(){
scanf("%s",ch);
len1=strlen(ch);
for(int i=len1-1;i>=0;i--)a[len1-i-1]=(cp){ch[i]-'0',0};
scanf("%s",ch);
len2=strlen(ch);
for(int i=len2-1;i>=0;i--)b[len2-i-1]=(cp){ch[i]-'0',0};
len=1;
while(len<(len1+len1)||len<(len2+len2))len<<=1;
for(int i=len1;i<len;i++)a[i]=(cp){0,0};
for(int i=len2;i<len;i++)b[i]=(cp){0,0};
fft(a,len,1);fft(b,len,1);
for(int i=0;i<len;i++)a[i]=a[i]*b[i];
fft(a,len,-1);
for(int i=0;i<len;i++)ans[i]=a[i].x+0.5;
for(int i=0;i<len;i++)ans[i+1]+=ans[i]/10,ans[i]%=10;
while(!ans[len-1]&&len)len--;
for(int i=len-1;i>=0;i--)printf("%d",ans[i]);
printf("\n");
return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <bitset>
#include <cmath>
#include <queue>
#include <stack>
using namespace std;

int main()
{
char a[10000] , b[10000];
int n[10000]={} , m[10000]={} , result[10000] = {};
int la , lb , i , j , w = 0;
int k = w;
cin >> a >> b;
la = strlen(a);
lb = strlen(b);
for(i = 0;i < la;i++) n[i] = a[la-i-1] - '0';
for(i = 0;i < lb;i++) m[i] = b[lb-i-1] - '0';
for(i = 0;i < la;i++)
for(j = 0;j < lb;j++)
{
result[i + j] += n[i] * m[j];
if(result[i + j] != 0) w = i + j;
}
for(i = 0;i <= w + 5;i++)
{
result[i + 1] += result[i] / 10;
result[i] %= 10;
if(result[i] != 0) k = i;
}
for(i = k;i >= 0;i--) cout << result[i];
return 0;
}

#include "stdio.h"
#include "string.h"

struct BigInt{
    int len;
    int nums[100000];
};

int main(){
    struct BigInt a, b, c;
    memset(&a, 0, sizeof(a));
    memset(&b, 0, sizeof(b));
    memset(&c, 0, sizeof(c));
    
    char input[100000];
    scanf("%s", input);
    for(int i = strlen(input) - 1; i >= 0 ; i--){
        a.nums[a.len++] = input[i] - 48;
    }
    
    scanf("%s", input);
    for(int i = strlen(input) - 1; i >= 0 ; i--){
        b.nums[b.len++] = input[i] - 48;
    }
    
    c.len = a.len + b.len;
    
    for(int i = 0; i < a.len; i++){
        for(int j = 0; j < b.len; j++){
            c.nums[i+j] = c.nums[i+j] + a.nums[i] * b.nums[j];
        }
    }
    
    for(int i = 0; i < c.len; i++){
        c.nums[i+1] = c.nums[i] / 10 + c.nums[i+1];
        c.nums[i] = c.nums[i] % 10;
    }
    
    while (c.nums[c.len] == 0 && c.len > 0){
        c.len--;
    }
    
    for(int i = c.len; i >= 0; i--){
        printf("%d", c.nums[i]);
    }
    
    printf("\n");
    
    return 0;
}
 

--Haskell一行

main = print . product . map read . words =<< getContents

裸题~

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <cstdlib>
using namespace std;

char str1[10002],str2[10002];
int a[10002],b[10002],c[20010];
int c1,c2;

void init()
{
    for(int i=0;i<c1;i++)
        a[c1-i-1]=str1[i]-'0';
    for(int i=0;i<c2;i++)
        b[c2-i-1]=str2[i]-'0';
}

int main()
{
    scanf("%s %s",str1,str2);
    c1=strlen(str1);
    c2=strlen(str2);
    init();
    for(int i=0;i<c1;i++)
        for(int j=0;j<c2;j++)
        {
            c[i+j]+=a[i]*b[j];
            if(c[i+j]>=10)
                c[i+j+1]+=c[i+j]/10,c[i+j]%=10;
        }
    int k;
    for(k=c1+c2;k>=0;k--)
        if(c[k]!=0) break;
    for(int i=k;i>=0;i--)
        cout<<c[i];
    return 0;
}
     

注意前面的0

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>

using namespace std;

int main()
{
string s1,s2;
cin>>s1>>s2;
int a[10001];
int b[10001];
int ans[100005];
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(ans,0,sizeof(ans));
a[0]=s1.length();
b[0]=s2.length();
for(int i=1;i<=a[0];i++)
{
a[i]=s1[a[0]-i]-'0';
}
for(int i=1;i<=b[0];i++)
{
b[i]=s2[b[0]-i]-'0';
}
int w,x=0;
w=a[0]+b[0];
for(int i=1;i<=a[0];i++)
{
x=0;
for(int j=1;j<=b[0];j++)
{
ans[i+j-1]=a[i]*b[j]+x+ans[i+j-1];
x=ans[i+j-1]/10;
ans[i+j-1]=ans[i+j-1]%10;
}
ans[i+b[0]]=x;
}
while(ans[w]==0&&w>1)
w--;
for(int i=w;i>=1;i--)
{
cout<<ans[i];
}
return 0;
}

#include <stdio.h>
#include <string.h>
char anss[110000000];
char * multipy(char * a, char * b)
{
int i,j,la,lb,l,k;
la=strlen(a);
lb=strlen(b);
l=la+lb;
char ans[l+1];
memset(ans,0,sizeof(ans));
//for(i=0;i<l;i++)printf("%d",ans[i]);
for (i=la-1;i>=0;i--)
{
//printf("i=%d\n",i);
for(j=lb-1;j>=0;j--)
{
//printf("j=%d\n",j);
k=l-la-lb+i+j+1;//printf("pre_ans[%d]=%d\n",k,ans[k]);
ans[k]+=(a[i]-'0')*(b[j]-'0');
ans[k-1]+=ans[k]/10;
ans[k]=ans[k]%10;
//printf("ans[%d]=%d\n",k,ans[k]);
}
}
k=0;
while(ans[k]!=0) k--;
if(ans[0]==0)i=1; else i=0;
for (j=i;j<l;j++) anss[j-i]=ans[j]+'0';//printf("%d",ans[i]);
return anss;
}
int main()
{
char input1[10001],input2[10001];
scanf("%s %s",input1,input2);
printf("%s",multipy(input1,input2));
return 0;
}

调了半天 才发现result开的10000+位
幡然醒悟应该开20000位
#include <cstdio>
#include <cstring>
#define Q 6000
#define clear(a) memset(a,0,Q*sizeof(int))

void input(int a[]){
clear(a);
char c[11000];
scanf("%s",c);
int len=1,k=1;
for(int i=strlen(c)-1;i>=0;i--){
if(k==10000)
k=1,len++;
a[len]+=k*(c[i]-'0');
k*=10;
}
a[0]=len;
}

void mult(int a[],int b[],int c[]){
clear(c);
for(int i=1;i<=a[0];i++)
for(int j=1;j<=b[0];j++){
c[i+j-1]+=a[i]*b[j];
c[i+j]+=c[i+j-1]/10000;
c[i+j-1]%=10000;
}
int len=a[0]+b[0];
while(len>1&&c[len]==0)len--;
c[0]=len;
}

void output(int a[]){
printf("%d",a[a[0]]);
for(int i=a[0]-1;i>=1;i--)
printf("%04d",a[i]);
}

int main(){
freopen("in.txt","r",stdin);
int x[Q],y[Q],r[Q];
input(x);
input(y);
mult(x,y,r);
output(r);
return 0;
}

#include<bits/stdc++.h>
using namespace std;
char A[50000000],B[5000000];
long long a[5000000],b[5000000],c[5000000];

int main()
{
int i,j,k=0,n,m,x;
long long l;
cin>>A>>B;
n=strlen(A); m=strlen(B);
strrev(A); strrev(B);

memset(a,0,sizeof(a)) ; memset(b,0,sizeof(b)) ; memset(c,0,sizeof(c)) ;
for(i=0;i<n;i++) a[i]=A[i]-48; for(i=0;i<m;i++) b[i]=B[i]-48;
k=0;
for (i=0;i<n;i++)
for (j=0;j<m+1;j++)
{
c[i+j] = c[i+j] + a[i] * b[j] +k;
k= c[i+j] / 10 ;
c[i+j] = c[i+j] % 10 ;
}
k=22000;
while (c[k]==0) k--;
for(i=k;i>=0;i--) cout<<c[i];
cout<<endl;
return 0;
}

某语言一行
print int(raw_input())*int(raw_input())