10! (factorial and factorial)

Well, this is not a factorial. It's the product of a term separated by another term. 10! = 10 * 8 * 6 * 4 * 2 = 3840
I learned high mathematics in University. It is estimated that the college entrance examination does not involve much

For example, 4x ^ 3,

In other words, 4x ^ 3 is the cube of 4 times X

Calculate 1! + 2! + 3!... + (n-1)! + n!. a C language program is designed to solve the problem. The calculation of factorial is realized by recursive function After running the program, input n, when n = 5, the output result is shown in the figure below

#include
#include
fun(int x);
int main(void)
{
char *pszBuff = NULL;
int a;
int i=1;
int sumResult = 0;
printf("input a int number:：");
scanf("%d", &a);
for(i=1;i

Please use the recursive call of the function to write the factorial function, and calculate 1! + 2! + 3! + 4! + 5!

#include
int fact(int);
main()
{
int i, sum=0;
for(i=1;i

In order to solve factorial problem, we need to define the function fact (n) to calculate n! In the main function, input a positive integer n, and output n! Require the function fact (n) to adopt recursion respectively To solve factorial problem, we need to define the function fact (n) to calculate n! And input a positive integer n in the main function and output n! Respectively. The function fact (n) is required to be implemented in recursive and non recursive forms

#include "stdio.h"
int fact1(int n)
{
if(n==1||n==0) return 1;
else return n*fact1(n-1);
}
int fact2(int n)
{
int i,t=1;
for(i=1;i

How many consecutive zeros are there in the result of factorial of 1000?

For every 2 and 5, there's a 0 at the end, so just look at it, from 1
The number of 2 and 5 in 1000 is enough. Because 5 is always less than 2, it is enough to see how many divisors 5 are in the factorial of 1000. Similarly, only the number with the end of 0 or 5 can have 5, so there are only 200 numbers including 5. However, there are 1000 / 25 = 40 numbers including 2 51000 / 125 = 8 numbers including 3 51000 / 625 = 1 number containing 4 5, So there are 200 + 40 + 8 + 1 = 249 5, so there are 249 consecutive 0

Factorial algorithm The factorial product of 36 divided by 35, what's the answer? What's the reason 34！ Why is it 1? 1 * 2 * 3 * 4 is greater than 1

The factorial of 36 is shown in this way
36!=1*2*3*…… *34*35*36
If you divide the above by 35, is 35 a contract? Then it becomes
1*2*3*…… *31*32*33*34*36
Write 36 alone, left 1 to 34 or complete, right? So it's the factorial of 34
So the answer is 36 (34!) (* can be omitted)

Recursive algorithm for finding factorial n

#includedouble fun(int n);int main(void){\x09int n;\x09printf("Enter n:");\x09scanf("%d",&n);\x09printf("%lf\n",fun(n));\x09return 0;}double fun(int n){\x09if(n==0||n==1)\x09\x09return 1;\x09else\x09\...

Factorial problem of n The problem is to take each number of factorials of N and do the prime factor decomposition. Finally, we can get the sum of the prime factors of the factorial of N. if n is 1 million, I will judge the number of 1 million, which is very troublesome, The radical n can be eliminated, and the following are prime numbers. But how do I do with the prime decomposition of those numbers that I reject? Isn't it a lot of numbers We should not only find out the prime number, but also divide the non prime number into the product of prime number multiplication, and finally output the sum of these prime numbers

No, it's looking

(n-1)! / (n-3)! How to do this factorial problem? Why (n-1)! Can be directly converted into (n-3)! (n-2) (n-1)

Because (n-1)! = 1 * 2 * 3 * 3 * 5 * 6 * *（n-5）*（n-4）*（n-3）*（n-2）*（n-1）
=(n-3)!*(n-2)*(n-1)

10! (factorial and factorial)