c++ 中的大整数

发布于 2024-08-13 01:41:25 字数 128 浏览 3 评论 0原文

我知道这个问题可能已经在这个论坛和网络上被问过很多次了。我被要求在 C++ 中创建一个大整数的实现，但是有一个限制，即我的构造函数之一应该采用 int 作为参数......所以我猜测会有多个非默认构造函数。所以我的问题是，最简单的方法是什么？

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執念 2024-08-20 01:41:25

那么，问题似乎是“如何将整数转换为位列表”？换句话说，整数的以 2 为底的表示形式是什么？

由于这应该是家庭作业，所以让我通过以 10 为基数的思考来讨论这个问题；经过一番思考，适当的改变应该是显而易见的。

给定一个以 10 为基数的数字，很容易算出最右边的数字是什么：它只是除以 10 时的余数。例如，如果 n=1234，那么它的最右边的数字是 n%10 = 4。要获取下一个最右边的数字，我们除以 10（得到 123），然后重复该过程。所以：

1234/10=123; 1234%10 = 4
123/10=12  ; 123%10 = 3
12/10=1    ; 12%10 = 2
1/10=0     ; 1%10 = 1

现在我们已经得到了答案[4,3,2,1]。如果我们反转它们，我们就会得到数字的基数 10 的数字：[1, 2, 3, 4]。

The question, then, seems to be "how do I turn an integer into a list of bits"? Put another way, what's the base-2 representation of an integer?

As this is supposed to be homework, let me talk around the problem by thinking in base-10; the appropriate changes should be obvious with some thought.

Given a base 10 number, it's pretty easy to figure out what the rightmost digit is: It's just the remainder when dividing by 10. E.g. if n=1234, then it's rightmost digit is n%10 = 4. To get the next rightmost digit, we divide by 10 (getting 123), and repeat the process. So:

1234/10=123; 1234%10 = 4
123/10=12  ; 123%10 = 3
12/10=1    ; 12%10 = 2
1/10=0     ; 1%10 = 1

So now we've gotten the answers [4,3,2,1]. If we reverse them, we have the base-10 digits of our number: [1, 2, 3, 4].

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指尖凝香 2024-08-20 01:41:25

C++ BigInt 类
 C++ 大整数库
例如编写 big int ：

typedef struct {
    int high, low;
} BiggerInt;

BiggerInt add( const BiggerInt *lhs, const BiggerInt *rhs ) {
    BiggerInt ret;

    /* Ideally, you'd want a better way to check for overflow conditions */
    if ( rhs->high < INT_MAX - lhs->high ) {
        /* With a variable-length (a real) BigInt, you'd allocate some more room here */
    }

    ret.high = lhs->high + rhs->high;

    if ( rhs->low < INT_MAX - lhs->low ) {
        /* No overflow */
        ret.low = lhs->low + rhs->low;
    }
    else {
        /* Overflow */
        ret.high += 1;
        ret.low = lhs->low - ( INT_MAX - rhs->low ); /* Right? */
    }

    return ret;
}

C++ BigInt class
C++ Big Integer Library
to write big int for example :

typedef struct {
    int high, low;
} BiggerInt;

BiggerInt add( const BiggerInt *lhs, const BiggerInt *rhs ) {
    BiggerInt ret;

    /* Ideally, you'd want a better way to check for overflow conditions */
    if ( rhs->high < INT_MAX - lhs->high ) {
        /* With a variable-length (a real) BigInt, you'd allocate some more room here */
    }

    ret.high = lhs->high + rhs->high;

    if ( rhs->low < INT_MAX - lhs->low ) {
        /* No overflow */
        ret.low = lhs->low + rhs->low;
    }
    else {
        /* Overflow */
        ret.high += 1;
        ret.low = lhs->low - ( INT_MAX - rhs->low ); /* Right? */
    }

    return ret;
}

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