我必須用 C 編程語言計算以下兩個數學公式。
- 結果 = 48^108 // 108 應該是 103
- 結果3 =9
要計算這兩個方程，我需要一些范圍足夠大的資料型別。
我知道等式 2 的最終結果是 9。但是，在使用了幾種資料型別后，我沒有得到這個答案。
這是我的一些實作。
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 | a3= pow(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 = 147179542864413390695231668723836254417826202083285489645297997883519171141486480221363409432872885235091123842885421688012169987663834748443552551569845821059256315786821632.000000
a4=46.000000
我應該使用哪種資料型別來處理這種情況？
我為我在方程 1 上的錯誤方程道歉。它是 103 而不是 108。如果我使用 103，答案是 9。
感謝您的所有意見和程序。
uj5u.com熱心網友回復：
48 103以 143 為模的余數很容易通過將指數分解為 2 的冪并保持所有中間值以 143 為模減少來計算，如以下代碼所示。（這不一定是使用最少算術運算的方法。）
#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);
}
uj5u.com熱心網友回復：
C 語言中沒有這樣的資料型別來存盤這么大的數字。您可以使用可以直接給出方程式答案的公式。
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