C ++中的双精度（或pow（2，1000）） [英] Double precision in C++ (or pow(2, 1000))

问题描述

我工作的项目欧拉刷上我的C ++编码技能的编程挑战（S），我们就做这个，下学期准备（因为他们不让我们使用Python，嘘！）。

I'm working on Project Euler to brush up on my C++ coding skills in preparation for the programming challenge(s) we'll be having this next semester (since they don't let us use Python, boo!).

我在＃16，我想找到一种方法来保持2¹°的真实精度。

I'm on #16, and I'm trying to find a way to keep real precision for 2¹°°°

例如：

int main(){
    double num = pow(2, 1000);
    printf("%.0f", num):
    return 0;
}

列印

10715086071862673209484250490600018105614050000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

10715086071862673209484250490600018105614050000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

一个缺少大部分的数字（从蟒蛇）：

Which is missing most of the numbers (from python):

>>> 2 ** 1000

10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376L

10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376L

授予，我可以用Python 1内核编写程序

Granted, I can write the program with a Python 1 liner

sum(int(_) for _ in str(2**1000))

立即给出结果，试图找到一种方法来做它在C ++。任何指针？ （haha ...）

that gives me the result immediately, but I'm trying to find a way to do it in C++. Any pointers? (haha...)

编辑：

标准libs之外的东西对我来说毫无价值 - - 树的代码是允许在那些比赛，而且我可能不会打印出10,000行的外部代码...

Something outside the standard libs is worthless to me - only dead-tree code is allowed in those contests, and I'm probably not going to print out 10,000 lines of external code...

推荐答案

如果你只是跟踪一个字符数组中的每个数字，这很容易。加倍数字是微不足道的，如果结果大于10，你只减去10和进位添加到下一位。从值1开始，循环加倍功能1000次，你就完成了。你可以预测你需要用 CEIL的位数（1000 *日志（2）/日志（10）），或者只是添加它们动态。

If you just keep track of each digit in a char array, this is easy. Doubling a digit is trivial, and if the result is greater than 10 you just subtract 10 and add a carry to the next digit. Start with a value of 1, loop over the doubling function 1000 times, and you're done. You can predict the number of digits you'll need with ceil(1000*log(2)/log(10)), or just add them dynamically.

垃圾邮件提醒：它显示我必须在任何人都相信我之前显示代码。这是一个带有两个函数Double和Display的bignum的简单实现。为了简单起见，我没有把它做成一个类。数字被存储在一个小尾数格式，用最少的显著位在先。

Spoiler alert: it appears I have to show the code before anyone will believe me. This is a simple implementation of a bignum with two functions, Double and Display. I didn't make it a class in the interest of simplicity. The digits are stored in a little-endian format, with the least significant digit first.

结果
结果
结果

typedef std::vector<char> bignum;

void Double(bignum & num)
{
    int carry = 0;
    for (bignum::iterator p = num.begin();  p != num.end();  ++p)
    {
        *p *= 2;
        *p += carry;
        carry = (*p >= 10);
        *p -= carry * 10;
    }
    if (carry != 0)
        num.push_back(carry);
}

void Display(bignum & num)
{
    for (bignum::reverse_iterator p = num.rbegin();  p != num.rend();  ++p)
        std::cout << static_cast<int>(*p);
}

int main(int argc, char* argv[])
{
    bignum num;
    num.push_back(1);
    for (int i = 0;  i < 1000;  ++i)
        Double(num);
    Display(num);
    std::cout << std::endl;
    return 0;
}

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