在 C 中寻找合适的数据类型来处理两个数学表达式

Question

我必须用 C 编程语言计算以下两个数学公式。

1. result = 48^108 // 108 应该是 103
2. result%143 =9

为了计算这两个方程，我需要一些范围足够大的数据类型。

我知道等式2的最终结果是9。但是，在使用了几种数据类型后我没有得到这个答案。

这是我的一些实现。

1. #include <stdio.h>
#include<math.h>
int main()
{
     unsigned int a1=48,a2=103,a3, a4;
    a3= pow(48,103);
    printf("a3=%u",a3);
    a4= a3 % 143;
    printf("\n a4=%u",a4);
    return 0;
}

 Answer That I got:
 warning: overflow in conversion from ‘double’ to ‘unsigned int’ changes value from ‘1.4717954286441339e+173’ to ‘4294967295’ [-Woverflow]
   15 |     a3= pow(48,103);
      |         ^~~
a3=4294967295
 a4=47

2. int main()
{
    unsigned long a1=48,a2=103,a3, a4;
    a3= pow(48,103);
    printf("a3=%lu",a3);
    a4= a3 % 143;
    printf("\n a4=%lu",a4);
    return 0;
}


warning: overflow in conversion from ‘double’ to ‘long unsigned int’ changes value from ‘1.4717954286441339e+173’ to ‘18446744073709551615’ [-Woverflow]

15 | 15 a3=战俘(48,103); | ^~~ a3=18446744073709551615 a4=16

3. #include <stdio.h>
#include<math.h>

int main()
{
    long double a1=48,a2=103,a3, a4;
    a3= pow(48,103);
    printf("a3=%LF",a3);
    a4= fmod(a3,143);
    printf("\n a4=%LF",a4);
    return 0;
}

a3=14717954286441339069523166872383625441782620208328548964529799788351917114148648022136340 9432872885235091123842885421688012169987663834748443552551569845821059256315786821632.000000

a4=46.000000

我应该使用哪种数据类型来处理这种情况？

我对方程 1 的错误方程表示歉意。它是 103，而不是 108。如果我使用 103，答案是 9。

感谢您的所有评论和过程。

Answer 1

通过将指数分解为 2 的幂并保持所有中间值减少模 143，可以轻松计算出 48¹⁰³ 模 143 的余数，如以下代码所示。 （这不一定是使用最少算术运算的方法。）

#include <stdio.h>


int main(void)
{
    //  Set x, y, and m for which we will compute x**y modulo m.
    unsigned x = 48, y = 103, m = 143;

    /*  r starts at the remainder of x**0 modulo m and is updated to be
        remainders of x raised to various powers as we compute them.
    */
    unsigned r = 1;

    /*  p will be the remainder of x modulo m, then of x**2 modulo m, then of
        x**4, x**8, x**16, and so on.  e will track this exponent, first, 1,
        then 2, then 4, 8, 16, and so on.

        For each bit that is set in y, say the bit representing 16, we will
        multiply r by the corresponding power (modulo m).  Thus, if y is 49 = 1
        + 16 + 32, we will multiply r by x**1, x**16, and x**32, all modulo m.
        This forms x**1 * x**16 * x**32 = x**(1+16+32) = x**49.  In general,
        the result will be x**y modulo m.

        e and p start at 1 and x, as described above.

        The loop continues as long as e is within the bits set in y, hence e <=
        y.

        In each iteration, we double e to move it to the next bit value
        position and we square p.  The squaring is done modulo m.

        Whenever e corresponds to a bit that is set in y ("y & e" is true), we
        multiply the current power in p by r and reduce the product modulo m.
     */
    for (unsigned e = 1, p = x; e <= y; e <<= 1, p = p*p % m)
        if (y & e)
            r = r*p % m;

    printf("%u\n", r);
}

Answer 2

C语言中没有这样的数据类型来存储如此庞大的数量。您可以使用可以直接给出方程式答案的公式。