这个东西有待考证和优化

代码如下

#include <iostream>
#include <cstdio>
#include <stack>
#include <cstring>
using namespace std;
//从这里到100行都是int128的运算代码
typedef struct _tword
{
    unsigned long long rah; //高64位
    unsigned long long ral; //低64位
    friend bool operator<(const _tword &a, const _tword &b)
    {
        if (a.rah < b.rah)
            return (1);
        if (a.rah > b.rah)
            return (0);
        return (a.ral < b.ral);
    }
} tword; //这个结构体就是128位整数了
unsigned long long hbt = 1;
unsigned long long man32;  //0~31位全1
unsigned long long man32h; //32~63位全1
void add(tword a, tword b, tword &res)
{
    unsigned long long al = a.ral & man32;
    unsigned long long ah = (a.ral & man32h) >> 32;
    unsigned long long bl = b.ral & man32;
    unsigned long long bh = (b.ral & man32h) >> 32;
    //因为这里需要处理进位，所以我们要把128位拆成4个32位并把它们扩充至64位，这样就能通过对结果的高32位的处理来加上进位
    al += bl;
    ah = ah + bh + ((al & man32h) >> 32);
    res.rah = a.rah + b.rah + ((ah & man32h) >> 32);
    res.ral = ((ah & man32) << 32) | (al & man32);
} //128位加法
void kuomul(unsigned long long a, unsigned long long b, tword &res)
{ //计算64位×64位=128位
    unsigned long long al = a & man32;
    unsigned long long ah = (a & man32h) >> 32;
    unsigned long long bl = b & man32;
    unsigned long long bh = (b & man32h) >> 32;
    //ah、al为a的高低32位，bh、bl为b的高低32位，则a*b=(ah*bh)<<32+(ah<<32)*bl+(bh<<32)*al+al*bl
    unsigned long long albl = al * bl;
    unsigned long long ahbh = ah * bh;
    unsigned long long albh = al * bh;
    unsigned long long ahbl = ah * bl;
    tword r1, r2, r3, r4;
    r1.rah = ahbh;
    r1.ral = 0;
    r2.rah = 0;
    r2.ral = albl;
    r3.ral = (albh & man32) << 32;
    r3.rah = (albh & man32h) >> 32;
    r4.ral = (ahbl & man32) << 32;
    r4.rah = (ahbl & man32h) >> 32;
    res.rah = 0;
    res.ral = 0;
    add(res, r1, res);
    add(res, r2, res);
    add(res, r3, res);
    add(res, r4, res); //把四项相加，得出结果
}
void mul(tword a, int b, tword &res)
{ //128乘int的运算
    tword tmp;
    unsigned long long ah = a.rah * b;
    kuomul(a.ral, b, tmp);
    unsigned long long al = tmp.ral;
    res.rah = ah + tmp.rah;
    res.ral = al;
}
int mod10(tword &hint)
{ //把128位hint除以10，并返回余数
    unsigned long long eah = hint.rah / 10;
    unsigned long long mod = ((hint.rah % 10) << 32) | ((hint.ral & man32h) >> 32);
    unsigned long long low = hint.ral & man32;
    hint.rah = eah;
    unsigned long long eal = (mod / 10) << 32;
    low = low | ((mod % 10) << 32);
    eal = eal | (low / 10);
    hint.ral = eal;
    return (low % 10);
}
stack<char> ss; //字符栈
void print(tword hint)
{ //输出128位整数
    if (hint.rah == 0 && hint.ral == 0)
    {
        printf("0"); //为0直接输出0
    }
    else
    {
        while (hint.rah != 0 || hint.ral != 0)
        {
            ss.push(mod10(hint) + '0');
        }
        while (!ss.empty())
        {
            putchar(ss.top());
            ss.pop();
        }
    }
}