大佬们看这个c语言有没有好办法增大它的指数范围

南山不会落梅花

//幂的计算
#include<stdio.h>
int main(void)
{
        printf("请输入底数:\n");
        float num;
        scanf_s("%f", &num);
        getchar();
        printf("请输入指数:(2≤指数≤4)\n");
        int exponent;
        scanf_s("%d", &exponent);
        getchar();

        if(exponent ==2)
        {
                float result = num * num;
                printf("结果是：%f", result);
        }
                if(exponent ==3)
        {
                float result = num * num * num;
                printf("结果是：%f", result);
        }
                if(exponent ==4)
        {
                float result = num * num * num * num;
                printf("结果是：%f", result);
        }

        getchar();
        return 0;
}

大佬们看这个c语言有没有好办法增大它的指数范围

YorkWaugh

1. #include <math.h>
   
2. #include <stdio.h>
   
3. int main()
   
4. {
   
5.     double a,x;
   
6.     printf("请输入底数:\n");
   
7.     scanf("%lf", &a);
   
8.     printf("请输入指数:\n");
   
9.     scanf("%lf", &x);
   
10.     printf("结果是：%lf\n", pow(a, x));
    
11.     return 0;
    
12. }

复制代码

不理解你的要求，这样可以吗

带刀侍卫

YorkWaugh 发表于 2022-11-25 20:24
不理解你的要求，这样可以吗

楼主想算底数=250，指数=250，结果是多少

os52

List of arbitrary-precision arithmetic software

Package-library name	Number type	Language	License
Boost Multiprecision Library	Integers, rationals, floats, and complex	C++ and backends using GMP/MPFR	Boost
TTMath	Integers, floats	C++	BSD
LibBF	Integers, floats	C	MIT
BeeNum	Integers, rationals	C++	MIT
longer-int	Integers	C	GPL
GNU Multi-Precision Library (and MPFR)	Integers, rationals, and floats	C and C++ with bindings	LGPL
CLN	Integers, rationals, floats, and complex	C++	GPL
ARPREC	Integers, floats, and complex	C++	BSD-type
MAPM, MAPM	Integers, decimal and complex floats	C (bindings for C++)	Freeware
MPIR (mathematics software)	Integers, rationals, and floats	C and C++ with bindings	LGPL
CORE	Integers, rationals, and floats	C++	Freeware
LEDA	Integers, rationals, and floats	C++	Freeware
CGAL	Integers, rationals, and floats	C++	LGPL
GeometricTools	Integers and rationals	C++	Boost
LibTomMath	Integers	C	Public Domain or WTFPL (dual-licensed)
libgcrypt	Integers	C	LGPL
OpenSSL	Integers	C	Apache License v2
Arbitraire	Floats	C	MIT License
mbed TLS	Integers	C	Apache License v2 and GPL
JScience	Integers, rationals, and floats	Java	BSD-type
JAS	Integers, rationals, and complex numbers	Java	LGPL
JLinAlg	Decimals, rationals, and complex numbers	Java	LGPL
Apfloat	Integers, rationals, floats, and complex numbers	Java, C++	LGPL
MPArith	Integers, rationals, floats, and complex numbers	Pascal, Delphi	Zlib
InfInt	Integers	C++	MPL
bigz	Integers, rationals	C (bindings for C++)	BSD-type
C++ BigInt Class	Integers	C++	GPL
ramp	Integers	Rust	Apache License v2
float	Floats	Rust	Apache License v2
fgmp	Integers	C	Public Domain
imath	Integers, rationals	ANSI C	MIT
hebimath	Integers, rationals, naturals, floats	C (C99)	MIT
bsdnt	Integers, naturals	C	BSD (2-clause)
integer-simple	Integers	Haskell	BSD (3-clause)
bigints	Integers	Nim	MIT
libzahl (WIP)	Integers	C	ISC
decimal	Decimals	Go	BSD (3-clause)
mpmath	Floats and complex	Python	BSD
Computable Reals	Computable Reals	Common Lisp	BSD (3-clause)
libmpdec and libmpdec++	Decimal floats	C and C++	BSD (2-clause)
GEM Library	Floats and complex numbers	MATLAB and GNU Octave	MPL

YorkWaugh

带刀侍卫 发表于 2022-11-25 21:18
楼主想算底数=250，指数=250，结果是多少

提供一个很丑陋的解决方案

1. #include <math.h>
   
2. #include <stdio.h>
   
3. #include <string.h>
   
4. int main()
   
5. {
   
6.     char a[10];
   
7.     int b = 0, x, d = 0, p[10000] = {}, q[10000] = {};
   
8.     printf("请输入底数:\n");
   
9.     gets(a);
   
10.     for (unsigned int i = 0; i < strlen(a); i++)
    
11.         b += (a[i] - 48) * pow(10, strlen(a) - i - 1);
    
12.     printf("请输入指数:\n");
    
13.     scanf("%d", &x);
    
14.     for (unsigned int i = 0; i < strlen(a); i++)
    
15.         p[strlen(a) - i - 1] = q[strlen(a) - i - 1] = a[i] - 48;
    
16.     for (int i = 1; i < x; i++)
    
17.     {
    
18.         for (int j = 1; j < b; j++)
    
19.             for (int k = 0; k < 10000; k++)
    
20.                 if (p[k] + q[k] < 10)
    
21.                     p[k] += q[k];
    
22.                 else
    
23.                     p[k + 1] += (p[k] + q[k]) / 10, p[k] = (p[k] + q[k]) % 10;
    
24.         for (int j = 0; j < 10000; j++)
    
25.             q[j] = p[j];
    
26.     }
    
27.     for (int i = 9999; i > 0; i--)
    
28.         if (p[i] != 0)
    
29.         {
    
30.             d = i;
    
31.             break;
    
32.         }
    
33.     printf("结果是：");
    
34.     for (int i = d; i >= 0; i--)
    
35.         printf("%d", p[i]);
    
36.     putchar('\n');
    
37.     return 0;
    
38. }

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结果为

1. 305493636349960468205197939321361769978940274057232666389361390928129162652472045770185723510801522825687515269359046715531785342780428396973513311420091788963072442053377285222203558881953188370081650866793017948791366338993705251636497892270212003524508209121908744820211960149463721109340307985507678283651836204093399373959982767701148986816406250000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

复制代码

thelord

改成循环，或者用数学库里的 pow函数（见2楼）

南山不会落梅花

YorkWaugh 发表于 2022-11-25 20:24
不理解你的要求，这样可以吗

感谢