Chapter 7 Experimental Functions
- Topic 1:
Trial quotient method is the simplest method to judge prime numbers. Use an integer between i=2 ~ m-1 to try the quotient. If there is a certain m that can be divisible by an integer i other than 1 and m itself (that is, the remainder is 0), then m is not a prime number. If all of the above ranges are If an integer cannot divide m, then m is a prime number. Using the trial quotient method, write the prime number judgment function IsPrime() with three methods: goto statement, break statement and set flag variables and strengthen the loop test. Input an integer m from the keyboard to determine whether m is a prime number, if m is a prime number , Then print the number as a prime number in the format "%d is a prime number\n", otherwise print the number as a prime number in the format "%d is not a prime number\n". Then analyze which method is more readable.
1. Problem solving ideas:
Define a flag, which is prime when flag=1. The prime number judges IsPrime(). If the number entered in the outer layer is 1 or a prime number, define prime=0 to return the main function, which is not a prime number. Trial quotient method, use for loop, i=2 ~ m-1 to try quotation, if the input number divided by the modulus of the i variable is 0, prime=0 returns the main function, this number is not a prime number, break.
2. Source code:
#include <stdio.h>
int m,prime;
int IsPrime(int m);
int main ()
{
int flag;
printf("please enter a number:",m);
scanf("%d",&m);
flag=IsPrime(m);
if(flag==1)
printf("%d is a prime number");
else
printf("%d is not a prime number",m);
return 0;
}
int IsPrime(int m)
{
if(m==1)
{
prime=0;
}
else
{
int i;
for(i = 2;i <= (m - 1);i++)
{
if(m%i==0)
{
prime=0;
break;
}
else
prime=1;
}
}
return prime;
}
3. Screenshot of program running effect:
- Topic 2:
Mathematical methods can be used to prove that a number that cannot be divisible by a number between 2 and √m (rounded) must not be divisible by 1 and any integer other than itself. According to this property of prime numbers, by modifying the specific realization of IsPrime(), programming to complete task 1.
- Problem-solving ideas:
On the basis of task 1, according to the nature of prime numbers, the range of variable i is changed to 2~√m.
2. Source code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int m,prime;
int IsPrime(int m);
int main ()
{
int flag;
printf("please enter a number:",m);
scanf("%d",&m);
flag=IsPrime(m);
if(flag==1)
printf("%d is a prime number");
else
printf("%d is not a prime number",m);
return 0;
}
int IsPrime(int m)
{
if(m==1)
{
prime=0;
}
else
{
int i;
for(i = 2;i <= sqrt(m);i++)
{
if(m%i==0)
{
prime=0;
break;
}
else
prime=1;
}
}
return prime;
}
3. Screenshot of program running effect:
- Question 3:
Input an integer n from the keyboard arbitrarily, program to calculate and output the sum of all prime numbers between 1 and n.
- Problem-solving ideas:
Define n, j, and 2 as prime numbers, and the sum of sum starting value is defined as 2. J starts from 3, if the return value of IsPrime(int j) is prime=1, it is a prime number, use a for loop, j=3~m, and sum accumulates j which is a prime number of 3~m and outputs it. For the prime number j, a loop is judged, whether the remainder of the variable i is 0, if it is not 0, return prime=1, and perform the accumulation calculation.
2. Source code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int m,prime,j;
int IsPrime(int j);
int main ()
{
int flag,sum=2;
printf("please enter a number:");
scanf("%d",&m);
for(j=3;j<=m;j++)
{
if(IsPrime(j))
{
sum += j;
}
}
printf("Sum is %d",sum);
return 0;
}
int IsPrime(int j)
{
int prime=1;
int i;
for(i=2;i<=j-1;i++)
{
if(j%i==0)
{
prime=0;
break;
}
else
prime=1;
}
return prime;
}
3. Screenshot of program running effect:
- Question 4:
Input an integer m from the health disk, if m is not a prime number, calculate and output all its factors (excluding 1), for example, for 16, output 2, 4, 8; otherwise, input "No divisor! It is a prime number" ".
1. Problem solving ideas:
Printf is placed in the for loop, and outputs the factors that are not prime numbers each time. For the input number is not a prime number, the outer program prime=0 returns to the main function, this number is the remainder of the variable j, and j loops once and adds 1 to it. If the remainder is 0, then j is its factor and output.
- Source code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int m,prime,j;
int IsPrime(int m);
int main ()
{
int flag,j;
printf("please enter a number:");
scanf("%d",&m);
flag=IsPrime(m);
if(flag==1)
{
printf("No divisor,It is a prime number");
}
else
{
for(j=2;j<m;j++)
{
if(m %j==0)
printf("%2d",j);
}
}
return 0;
}
int IsPrime(int m)
{
int prime=1;
int i;
for(i=2;i<=m-1;i++)
{
if(m%i==0)
{
prime=0;
break;
}
else
prime=1;
}
return prime;
}
3. Screenshot of program running effect:
- Question 5:
If all the different factors of a positive integer m that are less than m (including 1) add up exactly to m itself, then it is called a perfect number (Pefet Numberl) for a long time. For example, 6 is a perfect number, because 6=1+2+3, please write a function IsPerfect() that judges a perfect number, and then judge whether the integer input from the keyboard is a perfect number.
1. Problem solving ideas:
Change the function of IsPrime() to determine prime numbers to IsPerfect() to determine whether it is a perfect number. Sum defines the initial value of 0. For non-prime numbers, prime=0 returns to the main function in the outer layer. Variable j starts from 1. If the input number is 0 after the remainder of j is 0, it is a factor, and the factor j is accumulated. If all of m Different factors less than m (including 1) add up exactly to m, indicating that the input number is a perfect number.
- Source code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int m,prime,j;
int IsPerfect(int j);
int main ()
{
int flag,sum=0;
printf("please enter a number:");
scanf("%d",&m);
for(j=1;j<m;j++)
{
if(flag==1)
printf("");
else
{
if(m%j==0)
sum += j;
}
}
printf("Sum is %d",sum);
return 0;
}
int IsPerfect(int j)
{
int prime=1;
int i;
for(i=2;i<=j-1;i++)
{
if(j%i==0)
{
prime=0;
break;
}
else
prime=1;
}
return prime;
}
3. Screenshot of program running effect:
5. Test results of MOOC online courses
1. Test 1:
2. Test 2:
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