C语言 仅一个参数的质数递归函数
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C Recursive Function for prime number with just one parameter
提问by Rui Moreira
Im trying to make a function to check if a number is prime number or not, using recursion. The best two examples are these two programs (one without recursion, one using recursion).
我正在尝试使用递归来创建一个函数来检查一个数字是否为素数。最好的两个例子是这两个程序（一个没有递归，一个使用递归）。
Using recursion:
使用递归：
#include<stdio.h>
int isPrime(int,int);
int main(){
int num,prime;
printf("Enter a positive number: ");
scanf("%d",&num);
prime = isPrime(num,num/2);
if(prime==1)
printf("%d is a prime number",num);
else
printf("%d is not a prime number",num);
return 0;
}
int isPrime(int num,int i){
if(i==1){
return 1;
}else{
if(num%i==0)
return 0;
else
isPrime(num,i-1);
}
}
Without using recursion:
不使用递归：
#include<stdio.h>
int isPrime(int);
int main(){
int num,prime;
printf("Enter a positive number: ");
scanf("%d",&num);
prime = isPrime(num);
if(prime==1)
printf("%d is a prime number",num);
else
printf("%d is not a prime number",num);
return 0;
}
int isPrime(int num){
int i=2;
while(i<=num/2){
if(num%i==0)
return 0;
else
i++;
}
return 1;
}
My question is: Is it possible to make a function to check if a number is prime number or not, using recursion, but with just one parameter (like this: int isPrime(int num))?
我的问题是：是否可以使用递归来创建一个函数来检查一个数字是否为质数，但只有一个参数（例如：int isPrime(int num)）？
Any help is appreciated
任何帮助表示赞赏
回答by α?jiβ
You can use global variable
您可以使用全局变量
#include<stdio.h>
int isPrime(int);
int globalChk; //Global Variable
int main(){
int num=73;
int prime;
globalChk = num/2;
prime = isPrime(num);
if(prime==1)
printf("%d is a prime number",num);
else
printf("%d is not a prime number",num);
return 0;
}
int isPrime(int num){
if(globalChk==1){
return 1;
}else{
if(num%globalChk==0) {
return 0;
} else {
globalChk = globalChk-1;
isPrime(num);
}
}
}
回答by Simon Pham
float isPrime(float n) {
if ((int)n == 2)
return 1;
if ((int)(1/(n - (int)n)) % (int)n == 0)
return 0;
if (n / (int)n == 1)
return isPrime(n-1 + 1/n); // store original parameter to decimal places
return isPrime(n-1);
}
Usage:
用法：
if ((int)isPrime(n) == 1) {
// do this if n is a prime number
}
回答by ray lin
#include <stdio.h>
#include <math.h>
int isItPrime(double);
int main()
{
printf("Primes from 0 to 100:");
for (int n = 1; n < 100; n++)
{
if (isItPrime(n) == 1) //if it's a prime
printf(" %d", n);
}
return 0;
}
int isItPrime(double n) {
if(n<=1)
return 0;
long rounded = (long)n;//state of the iteration
double decimal = n - rounded;
if (decimal == 0){ //start condition for integers
return isItPrime( (double)(2.0 + 1.0/n)); //start with 2 then go all the way to sqrt(number being checked)
}
else {
long checkNum = round(1.0/decimal); //integer that is being checked for primality
if (checkNum == 2)
return 1;
long root = (long)(sqrt(checkNum));//square root of number being checked
if( rounded >= root) //exit condition
return !(checkNum % rounded == 0); //return prime if not a perfect square,
else{
if(checkNum % rounded == 0)
return 0;
else
return isItPrime((double)rounded + 1 + decimal); //recursive method call is here
}
}
}
Output: Primes from 0 to 100: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
输出：从 0 到 100 的素数：2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
回答by Manjunath N
Declare the variable num variable before main() function and pass the `num/2` to the isprime() function.
在 main() 函数之前声明变量 num 变量并将 `num/2` 传递给 isprime() 函数。
