求1到100之间所有整数的乘积的结果

像下面这种写法，sum的结果最终会严重超出int的范围，得不到想要的数据，求一个能显示正确结果的程序方法

#include <stdio.h>    
int main()
{
    
    int i=0;
    int sum=1;
    for(i=1;i<=100;i++)
    {
        sum=sum*i;
    }
    printf("%d",sum);
    return 0;
 } 

你需要实现一个任意精度整数乘法，下面是一个C++实现的例子，你可以参考一下。

#include <iostream>
#include <string>
#include <iterator>
#include <vector>

class BigInt
{
public:
    BigInt(unsigned int n = 0)
    {
        if (n == 0)
            _data.push_back('0');
        else
        {
            while (n)
            {
                _data.push_back(n % 10 + '0');
                n /= 10;
            }
        }
    }

    BigInt(const std::string &s) : _data(s.rbegin(), s.rend()) {}

    BigInt &operator+=(const BigInt &other);
    BigInt &operator*=(const BigInt &other);

    BigInt operator+(const BigInt &other) const
    {
        BigInt t = *this;
        t += other;
        return t;
    }

    BigInt operator*(const BigInt &other) const
    {
        BigInt t = *this;
        t *= other;
        return t;
    }

private:
    std::string _data;

    friend std::istream &operator>>(std::istream &is, BigInt &a);
    friend std::ostream &operator<<(std::ostream &os, const BigInt &a);
};

BigInt &BigInt::operator+=(const BigInt &other)
{
    int carry = 0;
    std::string::iterator itr1 = _data.begin();
    std::string::const_iterator itr2 = other._data.begin();
    while (itr1 != _data.end() && itr2 != other._data.end())
    {
        int a = *itr1 - '0';
        int b = *itr2 - '0';
        int c = a + b + carry;
        carry = c / 10;
        c %= 10;
        *itr1 = c + '0';
        ++itr1;
        ++itr2;
    }
    while (itr1 != _data.end())
    {
        int a = *itr1 - '0' + carry;
        carry = a / 10;
        a %= 10;
        *itr1 = a + '0';
        ++itr1;
    }
    while (itr2 != other._data.end())
    {
        int a = *itr2 - '0' + carry;
        carry = a / 10;
        a %= 10;
        _data.push_back(a + '0');
        ++itr2;
    }
    if (carry > 0)
        _data.push_back(carry + '0');
    return *this;
}

BigInt &BigInt::operator*=(const BigInt &other)
{
    BigInt r;
    for (std::size_t i = 0; i < other._data.size(); i++)
    {
        int x = other._data[i] - '0';
        int carry = 0;
        BigInt t = *this;
        for (std::string::iterator itr = t._data.begin(); itr != t._data.end(); ++itr)
        {
            int y = *itr - '0';
            int a = x * y + carry;
            *itr = a % 10 + '0';
            carry = a / 10;
        }
        if (carry > 0)
            t._data.push_back(carry + '0');
        if (i > 0)
        {
            std::string zeros(i, '0');
            t._data.insert(t._data.begin(), zeros.begin(), zeros.end());
        }
        r += t;
    }
    *this = r;
    return *this;
}

std::istream &operator>>(std::istream &is, BigInt &a)
{
    std::string s;
    is >> s;
    a = s;
    return is;
}

std::ostream &operator<<(std::ostream &os, const BigInt &a)
{
    std::copy(a._data.rbegin(), a._data.rend(), std::ostream_iterator<char>(os));
    return os;
}

int main()
{
    BigInt r = 1;
    for (int i = 1; i <= 100; i++)
        r *= i;
    std::cout << r << std::endl;
    return 0;
}

$ g++ -Wall main.cpp
$ ./a.out
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

构造个BigInt类，用数组来表示一个整数，用多位数运算基本规则实现BigInt类的乘法运算。

这个 用无符号64都不够。。。 越界啊！ 这个结果值太大了，如果你只是想了解计算机类型定义，int， unsigned int, long, longlong去梳理一下，这个结果太大了，完全超过了64位无符号能表示的数字的最大值，，，这就尴尬了，我是不知道计算机是否有方案设置更大的内存存对应数字的方案，，，

你可以打印看看输出，后面越界了。

int main()
{
    int i = 0;
    uint64_t sum = 1;
    for (i = 1; i <= 100; i++)
    {
        sum = sum * i;
        printf("%llu ", sum);
    }
    printf("%llu", sum);
    return 0;
}