topic:
Use trial division to print the prime numbers between 100 and 200 .
Prime number (prime number) : A number can only be written as the product of one and itself .
For example: 7 can only be written as 1*7 , that is a prime number (prime number).
=========================================================================
Idea 1: Use trial division
general idea:
(1). Use the outer loop : generate a number between 100 and 200.
(2). Set the inner loop : generate numbers from 2 to i-1 .
(3) . Set the if condition judgment statement in the inner loop ,
Check if i is a prime number :
Use a number between 2 and i-1 to try to divide i , if it can be divisible , then i is not a prime number ;
If no number between 2 and i-1 can divide i evenly , then i is a prime number .
(4). After the judgment is completed , judge whether it is a prime number according to the value of the variable flag , and print it out if it is .
first step:
(1). Use the outer loop : generate prime numbers between 100 and 200.
(2). Set a variable flag , if the flag is 1 , then i is a prime number , if the flag is 0 , then i is not a prime number .
Implementation code:
#include <stdio.h> int main() { int i = 0; //循环变量 for (i = 100; i <= 200; i++) //生成 100~200 之间的素数 { int flag = 1; //设置变量flag } return 0; }
Realize the picture:
Step two:
Set the inner loop : generate numbers from 2 to i-1.
Implementation code:
#include <stdio.h> int main() { int i = 0; //外循环变量 for (i = 100; i <= 200; i++) //生成 100~200 之间的素数 { int flag = 1; //设置变量flag int j = 0; //内循环变量 for (j = 2; j <= i - 1; j++) //设置内循环:生成 2~i-1 的数 { } } return 0; }
Realize the picture:
third step:
(1) . Set the if condition judgment statement in the inner loop .
(2). Determine whether i is a prime number :
Use a number between 2 and i-1 to try to divide i , if it can be divisible , then i is not a prime number ;
If no number between 2 and i-1 can divide i evenly , then i is a prime number .
Implementation code:
#include <stdio.h> int main() { int i = 0; //外循环变量 for (i = 100; i <= 200; i++) //生成 100~200 之间的素数 { int flag = 1; //设置变量flag int j = 0; //内循环变量 for (j = 2; j <= i - 1; j++) //设置内循环:生成 2~i-1 的数 { if (i % j == 0) //在内循环中设置 if条件判断语句,判断i是否为素数 //用 i 模上一个 j,看 j 能不能整除 i ,有余数则表示不能整除 { flag = 0; //flag == 0,则i不是素数 break; //只要有一个 j 把 i 整除了,说明 i 已经不是素数了, //所以不用再继续循环了,使用break跳出循环。 } } } return 0; }
Realize the picture:
the fourth step:
After the judgment , according to the value of the variable flag , it is judged whether i is a prime number , and if it is , it is printed out .
Implementation code:
#include <stdio.h> int main() { int i = 0; //外循环变量 for (i = 100; i <= 200; i++) //生成 100~200 之间的素数 { int flag = 1; //设置变量flag int j = 0; //内循环变量 for (j = 2; j <= i - 1; j++) //设置内循环:生成 2~i-1 的数 { if (i % j == 0) //在内循环中设置 if条件判断语句,判断i是否为素数 //用 i 模上一个 j,看 j 能不能整除 i ,有余数则表示不能整除 { flag = 0; //flag == 0,则i不是素数 break; //只要有一个 j 把 i 整除了,说明 i 已经不是素数了, //所以不用再继续循环了,使用break跳出循环。 } } if (flag == 1) //循环判断完后,根据变量的值,判断i是不是素数,是则打印 { printf("%d ", i); } } return 0; }
Realize the picture:
Idea 1: Final code and implementation effect
Final code:
#include <stdio.h> int main() { int i = 0; //外循环变量 for (i = 100; i <= 200; i++) //生成 100~200 之间的素数 { int flag = 1; //设置变量flag int j = 0; //内循环变量 for (j = 2; j <= i - 1; j++) //设置内循环:生成 2~i-1 的数 { if (i % j == 0) //在内循环中设置 if条件判断语句,判断i是否为素数 //用 i 模上一个 j,看 j 能不能整除 i ,有余数则表示不能整除 { flag = 0; //flag == 0,则i不是素数 break; //只要有一个 j 把 i 整除了,说明 i 已经不是素数了, //所以不用再继续循环了,使用break跳出循环。 } } if (flag == 1) //循环判断完后,根据变量的值,判断i是不是素数,是则打印 { printf("%d ", i); } } return 0; }
Realize the effect:
Summarize:
(1). Exercise the use of the outer loop and inner loop , and call the variables of the inner and outer loops .
(2). Trial division : use % to see the remainder and use it flexibly .
=========================================================================
Idea 2:
general idea:
Because even numbers are not prime except for 2 , and there is no 2 in the topic range ,
Therefore, only odd numbers between 100 and 200 can be generated , and half of the numbers can be excluded .
The efficiency is doubled .
first step:
Only on the basis of idea one ,
Just change the initialization part and adjustment part of the outer loop,
Make the outer loop only generate odd numbers between 100 and 200 .
Implementation code:
#include <stdio.h> int main() { int i = 0; //外循环变量 for (i = 101; i <= 200; i+=2) //生成 100~200 之间的奇数 { int flag = 1; //设置变量flag int j = 0; //内循环变量 for (j = 2; j <= i - 1; j++) //设置内循环:生成 2~i-1 的数 { if (i % j == 0) //在内循环中设置 if条件判断语句,判断i是否为素数 //用 i 模上一个 j,看 j 能不能整除 i ,有余数则表示不能整除 { flag = 0; //flag == 0,则i不是素数 break; //只要有一个 j 把 i 整除了,说明 i 已经不是素数了, //所以不用再继续循环了,使用break跳出循环。 } } if (flag == 1) //循环判断完后,根据变量的值,判断i是不是素数,是则打印 { printf("%d ", i); } } return 0; }
Realize the picture:
Idea 2: Final code and implementation effect
Final code:
#include <stdio.h> int main() { int i = 0; //外循环变量 for (i = 101; i <= 200; i+=2) //生成 100~200 之间的奇数 { int flag = 1; //设置变量flag int j = 0; //内循环变量 for (j = 2; j <= i - 1; j++) //设置内循环:生成 2~i-1 的数 { if (i % j == 0) //在内循环中设置 if条件判断语句,判断i是否为素数 //用 i 模上一个 j,看 j 能不能整除 i ,有余数则表示不能整除 { flag = 0; //flag == 0,则i不是素数 break; //只要有一个 j 把 i 整除了,说明 i 已经不是素数了, //所以不用再继续循环了,使用break跳出循环。 } } if (flag == 1) //循环判断完后,根据变量的值,判断i是不是素数,是则打印 { printf("%d ", i); } } return 0; }
Realize the effect:
=========================================================================
Idea three:
general idea:
a number: k ,
If k = m * n ,
Then there must be m or n less than root k ,
Then you can put the previous 2 ~ i-1
Change to sqrt(i) , that is, the root sign i ,
Further improve efficiency .
first step:
Only on the basis of the second idea ,
Just change the judgment condition part of the inner loop,
Reduce the number of executions of the inner loop .
Implementation code:
#include <stdio.h> #include <math.h> int main() { int i = 0; //外循环变量 for (i = 101; i <= 200; i += 2) //生成 100~200 之间的奇数 { int flag = 1; //设置变量flag int j = 0; //内循环变量 for (j = 2; j <= sqrt(i); j++) //设置内循环:生成 2~i-1 的数 { if (i % j == 0) //在内循环中设置 if条件判断语句,判断i是否为素数 //用 i 模上一个 j,看 j 能不能整除 i ,有余数则表示不能整除 { flag = 0; //flag == 0,则i不是素数 break; //只要有一个 j 把 i 整除了,说明 i 已经不是素数了, //所以不用再继续循环了,使用break跳出循环。 } } if (flag == 1) //循环判断完后,根据变量的值,判断i是不是素数,是则打印 { printf("%d ", i); } } return 0; }
Realize the picture:
Idea 3: Final code and implementation effect
Final code:
#include <stdio.h> #include <math.h> int main() { int i = 0; //外循环变量 for (i = 101; i <= 200; i += 2) //生成 100~200 之间的奇数 { int flag = 1; //设置变量flag int j = 0; //内循环变量 for (j = 2; j <= sqrt(i); j++) //设置内循环:生成 2~i-1 的数 { if (i % j == 0) //在内循环中设置 if条件判断语句,判断i是否为素数 //用 i 模上一个 j,看 j 能不能整除 i ,有余数则表示不能整除 { flag = 0; //flag == 0,则i不是素数 break; //只要有一个 j 把 i 整除了,说明 i 已经不是素数了, //所以不用再继续循环了,使用break跳出循环。 } } if (flag == 1) //循环判断完后,根据变量的值,判断i是不是素数,是则打印 { printf("%d ", i); } } return 0; }
Realize the effect: