Solve prime numbers between 100 and 200

solve 100~200 Between the prime number directory

List of articles

• solve 100~200 Between prime numbers Catalog
• Preface
• One 、 What are prime numbers ？
• Two 、 Method of solving prime numbers
  • 1. Conventional methods
  • 2. An optimization method
  • 3. To open or find a new path or snap course
• summary

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Preface

For many people, there is some confusion about solving prime numbers , ad locum , I will explain my understanding to you in some ways , I hope I can help you . Here is the solution 100~200 Between prime numbers

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One 、 What are prime numbers ？

prime number ： Prime numbers are also called prime numbers , A prime is defined as a number greater than 1 Of Natural number in , except 1 And there's no more than it itself Factor The natural number of .（PS：7=1*7,7 As a prime number ）

Two 、 Method of solving prime numbers

1. Conventional methods

The code is as follows （ Example ）：

#include<stdio.h>
// solve 100~200 The prime between 
int main()
{
	int i = 0;
	int j = 0;
	int sum = 0;                 //sum  Count , Statistics 100~200 The number of primes between   
	for (i = 100; i <= 200; i++) // Prescribed scope 
	{
		for (j = 2; j < i; j++)  // Prime numbers are defined as having only two factors ：1 And itself 
		{
			if (i % j == 0)      // If i%j==0  explain j It's also i Factor of , Not satisfying the condition of prime number 
			{
				break;           // Not satisfied , Out of the loop 
			}
		}
		if (j == i)              // If it's a prime number , Not in the internal circulation break operation ,j++ To the end of the cycle i
		{
			printf("%d ", i);    // Output i
			sum++;               // Add one to the total 
		}
	}
	printf("\nsum=%d\n", sum);
	return 0;
}

2. Optimize the code

Optimize the code 1 as follows （ Example ）：

int main()
{
	int i = 0;
	int j = 0;
	int sum = 0;                 //sum  Count , Statistics 100~200 The number of primes between   
	for (i = 101; i < 200; i+=2) //100,200 Even numbers are not considered ,100~200 Even numbers in do not consider each addition 2, skip 2
	{
		for (j = 2; j < i; j++)  // Prime numbers are defined as having only two factors ：1 And itself 
		{
			if (i % j == 0)      // If i%j==0  explain j It's also i Factor of , Not satisfying the condition of prime number 
			{
				break;           // Not satisfied , Out of the loop 
			}
		}
		if (j == i)              // If it's a prime number , Not in the internal circulation break operation ,j++ To the end of the cycle i
		{
			printf("%d ", i);    // Output i
			sum++;               // Add one to the total 
		}
	}
	printf("\nsum=%d\n", sum);
	return 0;
}

Through the understanding of the definition of prime numbers , Understand even numbers （ barring 0,2） It must not be a prime number , During the cycle , Even numbers are no longer considered , It can greatly improve the operation speed of the code .

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for (i = 101; i < 200; i+=2) //100,200 Even numbers are not considered ,100~200 Even numbers in do not consider each addition 2, skip 2
	{
		for (j = 2; j <= i/2; j++)  // Prime numbers are defined as having only two factors ：1 And itself ,i/2 For the previous session 
		{
			if (i % j == 0)      // If i%j==0  explain j It's also i Factor of , Not satisfying the condition of prime number 
			{
				break;           // Not satisfied , Out of the loop 
			}
		}
		if (j == i)              // If it's a prime number , Not in the internal circulation break operation ,j++ To the end of the cycle i
		{
			printf("%d ", i);    // Output i
			sum++;               // Add one to the total 
		}
	}

 

	for (i = 101; i < 200; i+=2) //100,200 Even numbers are not considered ,100~200 Even numbers in do not consider each addition 2, skip 2
	{
		for (j = 2; j <= sqrt(i); j++)  // Prime numbers are defined as having only two factors ：1 And itself ,
		{
			if (i % j == 0)      // If i%j==0  explain j It's also i Factor of , Not satisfying the condition of prime number 
			{
				break;           // Not satisfied , Out of the loop 
			}
		}
		if (j == i)              // If it's a prime number , Not in the internal circulation break operation ,j++ To the end of the cycle i
		{
			printf("%d ", i);    // Output i
			sum++;               // Add one to the total 
		}
	}

 

3. To open or find a new path or snap course

Using the method of function , Or adopt a special method , See the following （）：

#include<stdio.h>
int main()
{
	int i = 0;
	int sum = 0;                 //sum  Count , Statistics 100~200 The number of primes between   
	for (i = 101; i < 200; i+=2) //100,200 Even numbers are not considered ,100~200 Even numbers in do not consider each addition 2, skip 2
	{
		if (i % 2 != 0 && i % 3 != 0 && i % 5 != 0 && i % 7 != 0&&i%11!=0&&i%13!=0)  // because 2,3,5,7,11,13 Prime number 
		{
			printf("%d ", i);
			sum++;
		}
	}
	printf("\nsum=%d\n", sum);
	return 0;
}

stay 100~200 In the range of , If you can divide prime numbers by numbers , Explain that prime number is its factor , The reason for not considering other numbers is that other numbers have multiple factors , Prime numbers have only two factors , If you divide prime numbers by whole numbers, they are not equal to 0, Note that the number of factors included is only 1 And itself , Then it must be a prime number .

4. Function method to solve

#include<stdio.h>
#include<math.h>
int su(int a)             // You need to return a judgment number 
{
	int i = 0;
	for (i = 2; i <= sqrt(a); i++) 
	{
		if (a % i == 0)
		{
			return 1;     // Not prime numbers return 1
			break;        // End the cycle directly , End of the function 
		}
	}
}
int main()
{
	int i = 0;
	int sum = 0;
	for (i = 101; i < 200; i+=2)
	{
		if (su(i) != 1)
		{
			printf("%d ", i);
			sum++;
		}

	}
	printf("\nsum=%d\n",sum);
	return 0;
}

summary

In this subject , We need to understand the definition of prime numbers , Understand the judgment conditions , What are the closing conditions . Praise , Please pay attention ！