不知**高精度**怎么办?

HuangSiHan3116 @ 2024-07-18 22:38:38

#include<bits/stdc++.h>
using namespace std;
int main(){
    long long int a,b;
    cin>>a>>b;
    printf("%lld",a*b);
    return 0;
}

by jiangyelei @ 2024-07-23 16:38:17

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int q=1e4+4;
int main()
{
    char a1[q],b1[q];
    int a[q],b[q],c[q],lena,lenb,lenc,i,j,x;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    cin>>a1>>b1;
    lena=strlen(a1);
    lenb=strlen(b1);
    for(i=0;i<=lena-1;i++)
        a[lena-i]=a1[i]-48;
    for(i=0;i<=lenb-1;i++)
        b[lenb-i]=b1[i]-48;
    for(i=1;i<=lena;i++)
    {
        x=0;
        for(j=1;j<=lenb;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+lenb]=x;
    }
    lenc=lena+lenb;
    while (c[lenc]==0 && lenc>1)
        lenc--;
    for(i=lenc;i>=1;i--)
        cout<<c[i];
    cout<<'\n';
    return 0;
}
//c++

by haewa @ 2024-08-14 15:57:07

最简单的是加法吧。乘法还要双重循环 @wangcaizsr

by wangcaizsr @ 2024-08-14 17:18:27

@haewa 我是说用高精度做简单，不是说这个是高精度里最简单的