C language output you are prime Numbers all three digits, for advice-CodePudding

Won't do it, pure of new, how can not do a prime number

CodePudding user response:

Prime generally refers to prime, it is to point to in a natural number greater than 1, besides 1 and itself, there will be no other factor of natural Numbers, 2 hc-positie

Fyi:

 # include & lt; stdio.h> 

Int main (void) 
{
Int a [4]={2, 3, 5, 7}; 
Int I, j, k; 

for(i=0; i<4. I++) 
{
for(j=0; J<4. J++) 
{
for(k=0; K<4. K++) 
{
Printf (" % d % d % d \ n ", a [I], a [j], a [k]); 
} 
} 
} 

return 0; 
}

CodePudding user response:

From big to small, row shows the number 10,

 # include & lt; stdio.h> 

Int main (void) 
{
Int a [4]={2, 3, 5, 7}; 
Int I, j, k, c=0; 

For (I=3; I>=0; I -) 
{
For (j=3; J>=0; J -) 
{
For (k=3; K>=0; K -) 
{
C + +; 
Printf (" % d % d % d % c ", a [I], a [j], a [k], 10==0 c %? '\ n' : '\ t'); 
} 
} 
} 

return 0; 
}

CodePudding user response:

Bosses if from big to small how to judge

CodePudding user response:

refer to the second floor ctrigger response:

prime generally refers to prime, it is to point to in a natural number greater than 1, besides 1 and itself, there will be no other factor of natural Numbers, 2 hc-positie

Fyi:
 # include & lt; stdio.h> 

Int main (void) 
{
Int a [4]={2, 3, 5, 7}; 
Int I, j, k; 

for(i=0; i<4. I++) 
{
for(j=0; J<4. J++) 
{
for(k=0; K<4. K++) 
{
Printf (" % d % d % d \ n ", a [I], a [j], a [k]); 
} 
} 
} 

return 0; 
} 

You this just seems to be the three digits, printf (" % d \ n ", a [I] * 100 * 10 + a + a [j] [k]) ; So that the output to on the real of three digits,

CodePudding user response:

Topic mainly for judgment prime, plus a judgment function, for reference only.

 # include & lt; stdio.h> 

# define NUM 10 

/* determine whether prime */
Int is_prime (int n) 
{
Int I, the result; 

If (n==1) result=0; 
The else 
{
Result=1; 

For (I=2; i<=n/2; I++) 
{
If I (n %==0) 
{
Result=0; 
break; 
} 
} 
} 

return result; 
} 

Int main (void) 
{
Int a [NUM], len=0; 
Int I, j, k, c=0; 

/* to calculate an all primes, deposited in the array */
for(i=1; i<10; I++) 
{
If (is_prime (I)) 
{
A [len]=I; 
Len++; 
} 
} 

For (I=len - 1; I>=0; I -) 
{
For (j=len - 1; J>=0; J -) 
{
For (k=len - 1; K>=0; K -) 
{
C + +; 
Printf (" % d % d % d % c ", a [I], a [j], a [k], 10==0 c %? '\ n' : '\ t'); 
} 
} 
} 

return 0; 
}