C language small exercises

1. Print the prime numbers between 100-200

#include<stdio.h>
#include<Windows.h>
#include<math.h>

/*
  How to find prime numbers between 100 and 200
*/

int main() {

	
	//1. Intuitive method
	int i = 100;
	for (i = 100; i <= 200; i ++)
	{
		int j = 2;
		for (j = 2; j < i; j ++)
		{
			if (i % j == 0)
				break;
		}
		if (j >= i)
			printf("%d\n", i);
	}

	//2. Even numbers are multiples of 2, so they are not prime numbers, so we can exclude even numbers and directly determine whether odd numbers are prime numbers or not

	int i = 100;
	for (i = 100; i <= 200; i++)
	{
		int j = 2;
		for (j = 2; j < i / 2; j++)
		{
			if (i%j == 0)
				break;
		}
		if (j >= i / 2)
			printf("%d\n", i);
	}

	//3. Improvement method
	int i = 100;
	for (i = 101; i <= 200; i += 2)
	{
		int tem = sqrt(i);
		int j = 2;
		for (j = 2; j < tem; j ++)
		{
			if (i % j == 0)
				break;
		}
		if (j >= tem)
			printf("%d\n", i);
	}


	system("pause");
	return 0;
		
	
}

2. Ninety-nine multiplication formula table

#include<stdio.h>
#include<Windows.h>

int main() {

	int i, j; //define two loop variables
	for (i = 1; i < 9; i++) //outer loop
	{
		for (j = 1; j <= i; j++) //inner loop
		{
			printf("%d * %d = %d\t", i, j, i*j); //print formula
		}
		printf("\n"); //newline
	}

	system("pause");
	return 0;

}

3. Determine the leap year between 1000 and 2000

int test()
	{
		int count = 0;
		int year = 0;
		for (year = 1000; year <= 2000; year++)
		{
			if ((year % 4 == 0) && (year % 100 != 0) || (ear % 400 == 0))
			{
				printf("%d ", year);
				count++;
			}
		}
		printf("\ncount = %d\n", count);
		return 0;
	}