Template Collation ~~~~ Big Integer Multiplication

Large integer multiplication: Conceptually, it is a number that cannot be stored in int, long, long long, double, etc., and the length is several hundred digits

　　　　　　So we use strings to store and start from the least bit, and then the big integer multiplication turns into

　　　　　　The problem of vertical multiplication in elementary school.

 example:

Link: https://www.nowcoder.com/acm/contest/118/E
Source: Niuke.com

Topic description

Kirai likes to send "233" when chatting. "233" comes from Maopu emoji No. 233, which is a thumping and laughing emoji.

Kirai always replies "2333..." whenever she sees interesting news.
Kirai is actually very cold, he found this problem. In order not to want others to know that he was laughing immediately, he decided to multiply the two "233.." together and send it out.

Enter description:

There are multiple groups of input samples, all of which are positive integers. First, enter the number of sample groups T (T≤1500). 
Next, enter the number of T groups, each group of numbers consists of two 233 strings, each 233 string length 3≤n≤50. 
The data guarantees that each 233 string must have a 2 as the beginning, and the number of 3 is ≥ 2.

Output description:

The product of two 233 strings.

Example 1

enter

2
233 233
23333333333333333333333333333333333333333333333333 23333333333333333333333333333333333333333333333333

output

54289
544444444444444444444444444444444444444444444442888888888888888888888888888888888888888888888889 
_

#include<cstdio>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
char a[100],b[100];
int c[100],d[100];
void Bigintercheng(){
    memset(c,0,sizeof(c));
    memset(d,0,sizeof(d));
    int lena=strlen(a);
    int lenb=strlen(b);
    for(int i=0;i<lena;i++){
        c[i]=a[lena-i-1]-'0';
    }
    for(int j=lenb-1;j>=0;j--){
        for(int i=0;i<lena;i++){
            d[i+lenb-1-j]+=c[i]*(b[j]-'0');
        }
    }
    for(int i =0;i<lena+lenb-1;i++){
        d[i+1]+=d[i]/10;
        d[i]%=10;
    }
    int f=0;
    for(int i=500;i>=0;i--){
        if(!d[i]&&!f) continue;
        f=1;
        printf("%d",d[i]);
    }
    printf("\n");
}
intmain ()
{
    int t;
    scanf("%d",&t);
    while(t--){
        cin>>a>>b;
        Bigintercheng();
    }
}

View Code

The java code is as follows

Because java has a package for large integers, it is enough to solve it directly (java is good

import java.math.BigInteger;
import java.util.Scanner;
 
public  class Main {
 
    public static void main(String[] args) {
     
        int t;
        Scanner in = new Scanner(System.in);
        t=in.nextInt();
        for(int i=0;i<t;++i)
        {
            BigInteger a=in.nextBigInteger();
            BigInteger b=in.nextBigInteger();
            System.out.println(a.multiply(b));
        }
        in.close();
    }
 
}

View Code

The python code is as follows (life is too short, I use python

t = int(raw_input())
while t > 0:
    t-=1
    a, b = raw_input().split()
    print(int(a)*int(b))

View Code

Python Dafa is good!