Topic: Print prime numbers between 100 and 200
Prime number: That is, a prime number. In addition to 1 and itself, there are no other divisors, then the data is a prime number.
Method One: Trial Division
Defect: Data that exceeds half of i is definitely not a multiple of i. Many meaningless operations are performed above.
int main()
{
int i = 0;
int count = 0;
// 外层循环用来获取100~200之间的所有数据，100肯定不是素数，因此i从101开始
for(i=101; i<=200; i++)
{
//判断i是否为素数：用[2, i)之间的每个数据去被i除，只要有一个可以被整除，则不是素数
int j = 0;
for(j=2; j<i; j++)
{
if(i%j == 0)
{
break;
}
}
// 上述循环结束之后，如果j和i相等，说明[2, i）之间的所有数据都不能被i整除，则i为素数
if(j==i)
{
count++;
printf("%d ", i);
}
}
printf("\ncount = %d\n", count);
return 0;
}
Method 2: Every time you get a piece of data, you only need to check whether there is an element in the interval [2, i/2] that can be divisible by 2i, which means that i is not a prime number.
int main()
{
int i = 0;//
int count = 0;
for(i=101; i<=200; i++)
{
//判断i是否为素数
//2->i-1
int j = 0;
for(j=2; j<=i/2; j++)
{
if(i%j == 0)
{
break;
}
}
//...
if(j>i/2)
{
count++;
printf("%d ", i);
}
}
printf("\ncount = %d\n", count);
return 0;
}
Method three
/*
Method 2 still contains some repeated data, and is optimized again:
If i can be [2, sqrt(i)] If m is divisible by any integer between 2 ~ m-1, then i is not a prime number
Reason: If m can be divisible by any integer between 2 ~ m-1, one of its two factors must be less than or equal to sqrt( m), the other is greater than or equal to sqrt(m).
*/
int main()
{ int i = 0; int count = 0;
for(i=101; i<=200; i++)
{
//判断i是否为素数
//2->i-1
int j = 0;
for(j=2; j<=sqrt(i); j++)
{
if(i%j == 0)
{
break;
}
}
//...
if(j>sqrt(i))
{
count++;
printf("%d ", i);
}
}
printf("\ncount = %d\n", count);
return 0;
}
Method 4
Continue to optimize method three. As long as i is not divisible by any data between [2, sqrt(i)], then i is a prime number. However, in actual operation, i does not need to gradually increase from 101 to 200 because 2 and 3 are out of the equation. In addition, there will not be two consecutive adjacent data that are prime numbers at the same time.
int main()
{
int i = 0;
int count = 0;
for(i=101; i<=200; i+=2)
{
//判断i是否为素数
//2->i-1
int j = 0;
for(j=2; j<=sqrt(i); j++)
{
if(i%j == 0)
{
break;
}
}
//...
if(j>sqrt(i))
{
count++;
printf("%d ", i);
}
}
printf("\ncount = %d\n", count);
return 0;
}