问题
How can I find the factorial of a number (from 1 to 10) in C, without using:
- loop statements like for, while, and do while;
- conditional operators like if and case; and
- arithmetic operators like + , − , * , % , /, ++, −−?
FYI: I found this question in C aptitude.
回答1:
Since it is only 1 to 10, simply precompute it and store it in a simple int array of size 11. For the first element in the array put 1. It is not a valid input range for your problem but might as well be correct.
We need to store 11 elements instead of the 10 we need because otherwise we'd need to use operation "-" to get the right index. Subtraction is not allowed in your problem though.
int factorial(int x)
{
return precomputedArray[x];
}
回答2:
Here is a solution without loops, arithmetics, or conditionals and which does not resort to precomputation. It also does not use short-circuiting conditionals like && or || which are in practice equivalent to if. So this seems to be the first proper solution without any conditionals at all. Now in proper C without C++ features :)
#include <stdio.h>
#define uint unsigned int
void A(uint *a, uint *b)
{
uint tmp = *a & *b;
*a = (*a | *b) & ~tmp;
*b = tmp << 1;
}
#define REPEAT32(s) \
s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s
uint add(uint a, uint b)
{
REPEAT32(A(&a, &b);) return a;
}
uint bitexpand(uint b)
{
b = (b << 1) | b; b = (b << 2) | b; b = (b << 4) | b;
b = (b << 8) | b; b = (b << 16) | b;
return b;
}
void M(uint *acc, uint *a, uint *b)
{
*acc = add(*acc, *a & bitexpand(*b & 1));
*a <<= 1;
*b >>= 1;
}
uint mult(uint a, uint b)
{
uint acc = 0;
REPEAT32(M(&acc, &a, &b);) return acc;
}
uint factorial(int n)
{
uint k = 1;
uint result = 0;
result |= (bitexpand(n == 1) & k);
k = mult(k, 2); result |= (bitexpand(n == 2) & k);
k = mult(k, 3); result |= (bitexpand(n == 3) & k);
k = mult(k, 4); result |= (bitexpand(n == 4) & k);
k = mult(k, 5); result |= (bitexpand(n == 5) & k);
k = mult(k, 6); result |= (bitexpand(n == 6) & k);
k = mult(k, 7); result |= (bitexpand(n == 7) & k);
k = mult(k, 8); result |= (bitexpand(n == 8) & k);
k = mult(k, 9); result |= (bitexpand(n == 9) & k);
k = mult(k, 10); result |= (bitexpand(n == 10) & k);
return result;
}
int main(int argc, char **argv)
{
uint i;
/* Demonstration loop, not part of solution */
for (i = 1; i <= 10; i++)
{
printf("%d %d\n", i, factorial(i));
}
}
Updated: the discussion contained the claim that short-circuiting conditional like && would be acceptable in a solution that does not use if. Here is a simple macro that mimics two-way 'if' using && and obviously makes the whole problem much less interesting:
#define IF(i, t, e) \
(void)((i) && (goto then##__LINE__, 1)); goto else##__LINE__;
then##__LINE__: t; goto cont##__LINE__; \
else##__LINE__: e; cont##__LINE__: ((void)0);
You can then define
#define WHILE(c, s) \
loop##__LINE__: IF(c, s; goto loop##__LINE__, ((void)0)))
and then the rest of the problem becomes trivial.
回答3:
#include <stdio.h>
static const int factorial[] = {
1,
1,
2,
6,
24,
120,
720,
5040,
40320,
362880,
3628800,
};
/* Test/demo program. */
int main(void)
{
int i;
for (i = 0; i <= 10; ++i)
printf("%d %d\n", i, factorial[i]);
return 0;
}
(Anyone using this answer for a homework question either fails or has a teacher with a good sense of humor.)
(Bah, I was slow. Other people gave this answer already. Feel free to vote their answer up.)
回答4:
Maybe I'm solving someone's homework, but it looked like a fun challenge, anyways, here is my solution (compiles with warnings, but can't help those without making it look ugly(er))
EDIT: I have changed the program to make it support considerably longer factorials (up to 20 or so) and made the code a bit tidier by removing the lookup table inside prev().
#include <stdio.h>
#include <stdlib.h>
#define _if(CND, OP1, OP2) (((CND) && ((OP1) || 1)) || (OP2))
long long int add(long long int x, long long int y){
long long int r = x ^ y;
long long int c = x & y;
c = c << 1;
_if(c != 0, r = add(r, c), 1);
return r;
}
long long int prev(long long int x){
return add(x, -1);
}
long long int mult(long long int x, long long int y){
long long int r;
_if(x == 0,
r = 0,
_if(x == 1,
r = y,
r = add(y, mult(prev(x), y))));
return r;
}
long long int fac(long long int x){
long long int r;
_if(x < 2,
r = 1,
r = mult(x, fac(prev(x))));
return r;
}
int main(int argc, char**argv){
long long int i;
for(i = 0; i <= 20; i++)
printf("factorial(%lli) => %lli\n", i, fac(i));
return 0;
}
Sample run:
[dsm@localhost:~/code/c]$ gcc -o proc proc.c
[dsm@localhost:~/code/c]$ ./proc #/
factorial(0) => 1
factorial(1) => 1
factorial(2) => 2
factorial(3) => 6
factorial(4) => 24
factorial(5) => 120
factorial(6) => 720
factorial(7) => 5040
factorial(8) => 40320
factorial(9) => 362880
factorial(10) => 3628800
factorial(11) => 39916800
factorial(12) => 479001600
factorial(13) => 6227020800
factorial(14) => 87178291200
factorial(15) => 1307674368000
factorial(16) => 20922789888000
factorial(17) => 355687428096000
factorial(18) => 6402373705728000
factorial(19) => 121645100408832000
factorial(20) => 2432902008176640000
[dsm@localhost:~/code/c]$
回答5:
"+", "-" and "* " are explicitly prohibited, but "+=", "-=" and "*=" are not and so the recursive implementation becomes…
int factorial( int arg )
{
int argcopy = arg;
argcopy -= 1;
return arg == 1 ? arg : arg *= factorial( argcopy );
}
VC7 refuses to compile the above when in "compile as C source mode" – moans about the const L-value for "*=", but here is another variant of the same:
int factorial( int arg )
{
int argcopy1 = arg;
int argcopy2 = arg;
argcopy1 -= 1;
argcopy2 *= arg == 1 ? 1 : fact( argcopy1 );
return argcopy2;
}
回答6:
This is not a complete answer, but just different approaches to add() and mult() functions:
#define add(a, b) sizeof (struct { char x[a]; char y[b]; })
#define mult(a, b) sizeof (struct { char x[a][b]; })
(I believe that C, unlike C++, allows definition of new types inside a sizeof.)
Here is one more (totally nonportable) implementation of add() based on pointer arithmetic:
int add(int x, int y) {
return (int) &((char*) x)[y];
}
回答7:
Here is a solution (the only one so far) that actually solves the problem under the required limitations.
int fac( int n )
{
/* The is the binary representation of the function: */
/* 0000 => 0000000000000000001 */
/* 0001 => 0000000000000000001 */
/* 0010 => 0000000000000000010 */
/* 0011 => 0000000000000000110 */
/* 0100 => 0000000000000011000 */
/* 0101 => 0000000000001111000 */
/* 0110 => 0000000001011010000 */
/* 0111 => 0000001001110110000 */
/* 1000 => 0001001110110000000 */
/* 1001 => 1011000100110000000 */
int bit0 = n & 1;
int bit1 = (n & 2) >> 1;
int bit2 = (n & 4) >> 2;
int bit3 = (n & 8) >> 3;
int notbit0 = bit0 ^ 1;
int notbit1 = bit1 ^ 1;
int notbit2 = bit2 ^ 1;
int notbit3 = bit3 ^ 1;
return
(bit0 & notbit1 & notbit2 & bit3) << 18 |
(bit0 & notbit1 & notbit2 & bit3) << 16 |
(notbit1 & notbit2 & bit3) << 15 |
(notbit1 & notbit2 & bit3) << 11 |
(notbit1 & notbit2 & bit3) << 8 |
(notbit1 & notbit2 & bit3) << 7 |
(notbit0 & notbit1 & notbit2 & bit3) << 12 |
(notbit0 & notbit1 & notbit2 & bit3) << 10 |
(bit0 & bit1 & bit2 & notbit3) << 12 |
(bit1 & bit2 & notbit3) << 9 |
(bit0 & bit1 & bit2 & notbit3) << 8 |
(bit1 & bit2 & notbit3) << 7 |
(bit0 & bit2 & notbit3) << 5 |
(bit2 & notbit3) << 4 |
(notbit0 & bit1 & bit2 & notbit3) << 6 |
(bit0 & notbit1 & bit2 & notbit3) << 6 |
(notbit1 & bit2 & notbit3) << 3 |
(bit0 & bit1 & notbit2 & notbit3) << 2 |
(bit1 & notbit2 & notbit3) << 1 |
(notbit1 & notbit2 & notbit3);
}
Here is a test program:
#include <stdio.h>
int main()
{
int i, expected, j;
for( i = 0; i < 10; ++i )
{
expected = 1;
for( j = 2; j <= i; ++j )
{
expected *= j;
}
if( expected != fac( i ) )
{
printf( "FAILED: fac(%d) = %d, expected %d\n", i, fac( i ), expected );
}
}
}
回答8:
Use asm to write assembly code.
Or, precompile a program and execute it from your program.
Why would you impose such limits on your code?
回答9:
here is a solution that uses pointer arithmetics for arithmetics and function pointers for conditionals.
#include <stdio.h>
int fact(int n);
int mul(int a, int b)
{
struct s {
char _v[b];
};
struct s *p = (struct s*)0;
return (int) &p[a];
}
int add(int a, int b)
{
return (int) (&((char *)a)[b]);
}
int is_0(int n)
{
return (n == 0);
}
int fact_0(int n)
{
return 1;
}
int fact_n(int n)
{
return mul(n, fact(add(n,-1)));
}
int (*facts[2])(int) = {fact_n, fact_0};
int fact(int n)
{
return facts[is_0(n)](n);
}
int main(int argc, char **argv)
{
int i;
for(i = 0; i<=10; i++) {
printf("fact %d = %d\n", i, fact(i));
}
}
Sample Run:
~ > gcc -std=c99 fact.c
~ > ./a.out
fact 0 = 1
fact 1 = 1
fact 2 = 2
fact 3 = 6
fact 4 = 24
fact 5 = 120
fact 6 = 720
fact 7 = 5040
fact 8 = 40320
fact 9 = 362880
fact 10 = 3628800
回答10:
Produce a giant set of ternary operators returning a precalculated value for each allowed input. Use macros to compute the values.
回答11:
Calculating factorial is the first (and for many people, the last) time you'll use recursion. The standard implementation is
long fact(int x)
{
if (x < 2)
return 1L;
else
return fact(x - 1) * x;
}
Some would argue that that last statement should be "x * fact(x-1)" so that the compiler can recognize that it's tail recursion. Personally, I doubt any compiler is smart enough to see it in that form and not see it in the other form.
However, since you've restricted it to not use "if" or "-", I don't know how you'd do it.
回答12:
rough sketch (already proposed by others!)
int[] factorials = {1,1,2,6,24, 120,720, ..etc };
return factorials[i];
回答13:
i too tried by putting the values in array. here i have used if conditions and while loops but no arithmetic operators involved.! trying if i could remove them too.
#include <stdio.h>
int add(int a, int b)
{
int t1, t2, ab, bb, cb=0, orb=1, ans=0;
do {
t1 = a >> 1;
t2 = t1 << 1;
if (a==t2) ab=0; else ab=1;
t1 = b >> 1;
t2 = t1 << 1;
if (b==t2) bb=0; else bb=1;
if (ab==1 && bb==1) {
if (cb==1) ans=ans | orb;
cb = 1;
}
if ( ab!=bb ) {
if (cb==0) ans = ans | orb;
}
if (ab==0 && bb==0) {
if (cb==1) {
ans = ans | orb;
cb=0;
}
}
orb = orb << 1;
a = a >> 1;
b = b >> 1;
} while (a!=0 || b!=0);
if (cb==1) ans = ans | orb;
return ans;
}
int multiply(int x,int y)
{
int result = 0, i = 0 , j=0;
while((i=add(i,1)) <= y)
result = add(result,x);
return result;
}
int factorial(int x)
{
if(x==1)
return 1;
else
return multiply(x,factorial(x-1));
}
int main()
{
int x;
printf("Enter a number between 0 and 10: ");
scanf("%d" , &x);
printf("\nFactorial: %d\n" , factorial(x));
return 0;
}
回答14:
Let's see if we can do something half-elegant, without depending on 1 <= n <= 10.
- Instead of looping we'll of course use recursion.
- Instead of an if for terminating the recursion, we'll use an array of function pointers!
(We still need comparison operators, such as<and==.)
EDIT: damaru used the function pointers trick first.
This gives: [All code is untested, no C compiler under hand!]
typedef int (*unary_fptr)(int);
int ret_1(int n) {
return 1;
}
int fact(int n) {
unary_fptr ret_1_or_fact[] = {ret_1, fact};
return multiply(ret_1_or_fact[n > 1](sub_1(n)), n);
}
We still need to implement sub_1 and multiply. Let's start with sub_1, which is a simple recursion on the bits until the carry stops (if you don't understand this, the similar add_1 at the end is simpler to think about):
int identity(int n) {
return n;
}
int sub_1(int n) {
unary_fptr sub_1_or_identity[] = {sub_1, identity};
int lsb = n & 1;
int rest = sub_1_or_identity[lsb](n >> 1);
return (rest << 1) | (lsb ^ 1);
}
multiply: The simplest I can think of is Russian Peasant multiplication, which reduces it to binary shifts and addition. With conditionals, a recursive formulation would look like this:
/* If we could use conditionals */
int multiply(int a, int b) {
int subproduct;
if(a <= 1) {
subproduct = 0;
} else {
subproduct = multiply(a >> 1, b << 1);
}
if(a & 1) {
return add(b, subproduct);
} else {
return subproduct;
}
}
Without conditionals, we have to use the dispatch array trick twice:
typedef int (*binary_fptr)(int, int);
int ret_0(int a, int b) {
return 0;
}
int multiply(int a, int b) {
binary_fptr ret_0_or_multiply = {ret_0, multiply};
int subproduct = ret_0_or_multiply[a >= 2](a >> 1, b << 1);
binary_fptr ret_0_or_add = {ret_0, add};
return ret_0_or_add[a & 1](subproduct, b);
}
Now all we miss is add. You should by now guess how it will go - a simultaneous recursion over bits of the two numbers, which reduces the problem to shifts and add_1:
int add(int a, int b) {
int lsb = (a & 1) ^ (b & 1);
int carry = (a & 1) & (b & 1);
binary_fptr ret_0_or_add = {ret_0, add};
int subsum = ret_0_or_add[(a >= 2) & (b >= 2)](a >> 1, b>> 1);
unary_fptr identity_or_add_1 = {identity, add_1};
return identity_or_add_1[carry](subsum << 1);
}
and add_1 is a simple recursion over bits until the carry stops:
int add_1(int n) {
unary_fptr identity_or_add_1[] = {identity, add_1};
int lsb = n & 1;
int rest = identity_or_add_1[lsb](n >> 1);
return (rest << 1) | (lsb ^ 1);
}
That's it I think! [As noted above all code is untested!]
回答15:
If you cannot use recursion, or arithmetic and you have a limited range of inputs, you could hard-code the result to be an array lookup,
so:
return factorials[x];
where you've pre-filled factorials with the relevant values
回答16:
what if the we have to valculate factorials of 1 to 100. How to store this big numbers?
回答17:
#include<stdio.h>
void main()
{
unsigned long int num,fact,counter;
while(counter<=num)
{
printf("Enter the number");
scanf("%d",&num);
fact=fact*counter;
counter++;
printf("The factorial of number entered is %lu",fact);
}
printf("press any key to exit...");
getch();
}
回答18:
Since it didn't say not to use library functions:
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
int main( int argc, char** argv)
{
printf( "%d\n", (int)round( exp( lgamma(2))));
printf( "%d\n", (int)round( exp( lgamma(3))));
printf( "%d\n", (int)round( exp( lgamma(4))));
printf( "%d\n", (int)round( exp( lgamma(5))));
printf( "%d\n", (int)round( exp( lgamma(6))));
printf( "%d\n", (int)round( exp( lgamma(7))));
printf( "%d\n", (int)round( exp( lgamma(8))));
printf( "%d\n", (int)round( exp( lgamma(9))));
printf( "%d\n", (int)round( exp( lgamma(10))));
printf( "%d\n", (int)round( exp( lgamma(11))));
return 0;
}
来源：https://stackoverflow.com/questions/653961/factorial-in-c-without-conditionals-loops-and-arithmetic-operators