C language test question 3 (program multiple choice question - including detailed explanation of knowledge points)

Basic input and output and process control

1.
#include <stdio.h>
main()
{
     int a=1,b=3,c=5;
if (c==a+b)
 printf("yes\n");
else
 printf("no\n");
}
 The running result is ：no

 See the textbook for details p89  Selection structure 
 See the textbook for details p91 Relational symbols 
 See Appendix for details D p378 The priority of the symbol 
== Indicates whether the values on both sides of the symbol are equal ;= Indicates that the value to the right of the symbol is assigned to the variable on the left 
 The test point of this question is choice structure 3 The second of the basic forms 

 Choose one of the three general forms of structure “ sentence ” They are compound sentences , Compound sentences should use {
     } Cover up , Only when a compound statement contains only one statement, it can be omitted {
     }, That's how it works , So two printf The operation did not add {
     }

 if c==a+b establish , execute printf("yes\n");
 otherwise （ namely c==a+b Don't set up ）, perform printf("no\n");

+ Has a higher priority than ==, So let's start with a+b, The value is 4,  expression 5==4 Don't set up , So execute printf("no\n"); That is, the output string no

2.  
#include <stdio.h>
main()
{
     int a=12, b= -34, c=56, min=0;
min=a;
  if(min>b)  
min=b;
  if(min>c)  
min=c;
printf("min=%d", min);
}
 The running result is ： min=-34
 See the textbook for details p89  Selection structure 
 The test point of this question is choice structure 3 The first of the three basic forms 
 There are two selection structures （ Two if sentence ）
	 Defining variables , And the assignment   here a=12, b= -34, c=56, min=0
	 take a Median copy , Assign to min, covers min Medium 0, here min The value in is updated to 12.
 if min>b establish , execute min=b;
 if min>c establish , execute min=c;
 Output min The value in 

12 Greater than -34, first if The expression of the statement holds , So execute min=b;  After execution min The value in is updated to -34.
-34 Less than 56, the second if The expression of the statement does not hold , So do not execute min=c;

 The final output min The value in , by -34.



3.
#include <stdio.h>
main()
 {
     int x=2,y= -1,z=5;
   if(x<y)
if(y<0) 
z=0;
else
 z=z+1;
  printf(“%d\n”,z);
 }
 The running result is ：5
 Encountered selection structure , First of all, we should clarify what operations to perform when the conditional expression is true . In this question , first if sentence , The following compound statement has no braces {
     }, Note that the compound statement contains only one statement , And then omit {
     }. The inner layer of the if...else... Is the second basic form of selection structure , Structurally, it is regarded as a statement . So the inner if...else... As the first if A compound statement of a statement .

 If the expression x<y establish , Then continue to judge 
           if y<0, execute z=0;
 otherwise （ namely y>=0）, perform z=z+1;
 Output z
2>-1, expression x<y Don't set up , Therefore, the inner if…else…. , in turn, z The value in has not been changed .
 Output z The value of 5

4. 
#include <stdio.h>
main()
{
     float a,b,c,t;
a=3;
b=7;
c=1;
if(a>b)
   {
    t=a;a=b;b=t;}
if(a>c)
   {
    t=a;a=c;c=t;}
if(b>c)
    {
    t=b;b=c;c=t;}
printf("%5.2f,%5.2f,%5.2f",a,b,c);
}
    The running result is ：1.00,  3.00,  7.00

 See the textbook for details p72 The output form of data 
 This question contains 3 individual if sentence , Every if After statement {
     } Can not be omitted , Because of every {
     } Contains multiple statements 
 If the expression a>b establish , execute {
    t=a;a=b;b=t;}
 If the expression a>c establish , execute {
    t=a;a=c;c=t;}
 If the expression b>c establish , execute {
    t=b;b=c;c=t;}
 Output a,b,c The value in , The width of each data output is required to be 5 A space , The decimal part is reserved 2 position , Data right alignment 

3 Less than 7, So the expression a>b Don't set up , So do not execute {
    t=a;a=b;b=t;}

3 Greater than 1, So the expression a>c establish , execute {
    t=a;a=b;b=t;}. The first sentence , take a Medium 3 Copy , Paste the t in ; The second sentence , take c Medium 1 Copy , Paste the a in , Overwrite the previous 3; The third sentence . take t Medium 3 copy to c in , overwrite c Previous in 1.  After executing the compound statement, it realizes a,c Exchange of values of elements ,a by 1,c by 3,t by 3,.

7 Greater than c Medium 3, therefore b>c establish , Execute, execute {
    t=b;b=c;c=t;}, The process is the same as above , After execution b by 3,c by 7,t by 7

 Now output a,b,c The value of 1.00,  3.00,  7.00

5．
#include < stdio .h >
main ( )
{
      float  c=3.0 , d=4.0;
if ( c>d ) c=5.0;
else 
if ( c==d ) c=6.0;
else  c=7.0;
 printf ( “%.1f\n”,c ) ;
}
 The running result is ：7.0
 The title is if...else... Nesting of statements , second if...else... As the first if...else... sentence else Partial compound statements .

 If the expression c>d establish , execute c=5.0;
 otherwise （ expression c>d Don't set up ）
			 If the expression c==d establish , execute c=6.0;
			 otherwise , perform c=7.0;
 Output c The value in 

3.0 Less than 4.0, So the expression c>d Don't set up , Execute the second if…else….
3.0 It's not equal to 4.0, So the expression c==d Don't set up , perform c=7.0, take 7.0 Assign to c,  overwrite c Medium 3.0, here c The value of 7.0
 Output at this time c The value in 

6.
#include <stdio.h>
main()
{
    	int m;
	scanf("%d", &m);	
	if (m >= 0)				0
	    {
    	if (m%2 == 0)	printf("%d is a positive even\n", m);
		    else		printf("%d is a positive odd\n", m);	}
	else 
 {
    	if (m % 2 == 0)  printf("%d is a negative even\n", m);	
		else	  printf("%d is a negative odd\n", m);	}
}
 If you type －9, Then the running result is :   -9 is a negative odd


7. 
#include <stdio.h>
main()
{
     int num=0;
while(num<=2){
     num++;printf("%d\n",num);}
}
 The running result is ：
1
2
3

 See the textbook for details p115 Loop structure 
 When the cyclic condition num<=2 When it was founded , Execution loop body {
     num++;printf("%d\n",num);} The statement in .

 The initial value of the cycle num by 0;
 The loop condition num<=2 establish 
 The first 1 Secondary cycle ： perform num++; the num Add... To the value in 1, After execution num by 1;
			 perform printf("%d\n",num); Output... On the screen num The value in , The output 1, After that, change the line 

 here num The value of 1, The loop condition num<=2 establish 
 The first 2 This cycle ： perform num++; the num Add... To the value in 1, After execution num by 2;
			 perform printf("%d\n",num); Output... On the screen num The value in , The output 2, After that, change the line 

 here num The value of 2, The loop condition num<=2 establish 
 The first 3 This cycle ： perform num++; the num Add... To the value in 1, After execution num by 3;
			 perform printf("%d\n",num); Output... On the screen num The value in , The output 3, After that, change the line 
 here num The value of 3, The loop condition num<=2 Don't set up , End of cycle .

8．
#include <stdio.h>
main( )
{
     int sum=10,n=1;
while(n<3)   {
    sum=sum-n;  n++; } 
printf(“%d,%d”,n,sum); 
} 
 The running result is ：3,7

 When the cyclic condition n<3 When it was founded , Execution loop body {
    sum=sum-n;  n++; } The statement in .

 The initial value of the cycle sum by 10,n by 1;
 The loop condition n<3 establish 
 The first 1 Secondary cycle ： perform sum=sum-n=10-1=9;
			 perform n++, the n Add... To the value in 1, After execution n by 2;

 here n The value of 2,sum The value of 9, The loop condition n<3 establish , Continue the loop 
 The first 2 Secondary cycle ： perform sum=sum-n=9-2=7;
			 perform n++, the n Add... To the value in 1, After execution n by 3;

 Output at this time n,sum The value in , That is to say 3,7. We need to pay attention to , stay printf(“%d,%d”,n,sum);  The output data required in are separated by commas , Therefore, there must be a comma between the two data of the result 




9.
#include <stdio.h>
main()
{
     int num,c;
 scanf("%d",&num);
 do  {
    c=num%10;  printf("%d",c);  }while((num/=10)>0);
printf("\n"); 
}
 Input from keyboard 23, Then the running result is ：32
 See the textbook for details p117 Loop structure ;p60 Composite assignment operators 
do{
     }while( expression );
 Execute the loop body unconditionally first , And then judge the cyclic condition . Be careful while（ expression ） It's followed by a semicolon 

 Defining integer variables num,c;
 by num Assign an integer value ;
 perform {
    c=num%10;  printf("%d",c);  } Until the cycle conditions (num/=10)>0 Don't set up ;
 Output line feed 

 Known as num assignment 23
 The first 1 Secondary execution loop body 
 perform c=num%10=23%10=3;
 perform printf("%d",c); Output 3
 Judge cyclic condition  num/=10 Equivalent to num=num/10;  therefore num=23/10=2, 2 Greater than 0, So the cyclic conditions (num/=10)>0 establish , Continue to execute the loop body . Finish the 1 In the second cycle ,num by 2,c by 3


 The first 2 Secondary execution loop body 
 perform c=2%10=2;
 perform printf("%d",c); Then the output 2
 Judge cyclic condition num=2/10=0,0 be equal to 0, So the cyclic conditions (num/=10)>0 Don't set up . End of cycle 

10
#include <stdio.h>
main()
{
     int s=0,a=5,n;
scanf("%d",&n);
do   {
     s+=1;  a=a-2; }while(a!=n);
printf("%d,%d\n",s,a);
 }
 If the value entered 1, The running result is ： 2,1 
 See the textbook for details p117 Loop structure ;p60 Composite assignment operators 
 perform {
     s+=1;  a=a-2; } Until the cycle conditions a!=n Don't set up ;

 Known as n assignment 1,s by 0,a by 5
 The first 1 Secondary execution loop body 
 perform s+=1; Equivalent to s=s+1=0+1
 perform a=a-2; a=5-2=3
 Judge cyclic condition ,3 It's not equal to 1, So the cyclic conditions a!=n establish , Continue to execute the loop body .
 Finish the 1 In the second cycle ,s by 1,a by 3
 The first 2 Secondary execution loop body 
 perform s+=1; Equivalent to s=s+1=1+1=2
 perform a=a-2; a=3-2=1
 Judge cyclic condition ,1 be equal to 1, So the cyclic conditions a!=n Don't set up , End of cycle .
 Finish the 2 In the second cycle ,s by 2,a by 1

 Output at this time s,a The value in , The result is 2,1

11．
#include "stdio.h" 
main()                                 
{
    char c;                               
c=getchar();                           
while(c!='?')     {
    putchar(c);    c=getchar(); }                            
}     
 If you type from the keyboard abcde？fgh（ enter ）
 The running result is ：abcde

12．
#include <stdio.h>
main()
{
     char c;
  while((c=getchar())!=’$’)
        {
     if(‘A’<=c&&c<=‘Z’)  putchar(c);
          else if(‘a’<=c&&c<=‘z’)  putchar(c-32);  }
} 
 When the input is ab*AB%cd#CD$ when , The running result is ：ABABCDCD

13.  
#include <stdio.h>
main()
{
     int x, y =0;
for(x=1;x<=10;x++)
{
     if(y>=10) 
 break; 
y=y+x;
 } 
printf(“%d   %d”,y,x);
}
 The running result is ：10   5
 See the textbook for details p120  for sentence 
 See the textbook for details p126~128  break,continue sentence 
for(  expression 1; expression 2; expression 3)
{
    

}
(1)  So let's solve for the expression 1 
(2)  Solution expression 2, If the value is true , Execution loop body , And then execute  (3).  If it's false , End cycle , go to (5)
(3)  Solution expression 3 
(4)  Turn back up  (2) Carry on 
(5)  The loop ends , perform for A statement below the statement  

break , Jump out of the loop ;continue, End this cycle （ The first i Secondary cycle ）, Continue with next cycle ( The first i+1 Secondary cycle )

 This question   expression 1 by x=1, expression 2（ The loop condition ） by x<=10, expression 3 by x++
 initial value x by 1,y by 0, The loop condition （ That's the expression 2）x<=10 establish , Into the circulatory body 

 The first 1 Secondary cycle 
	 perform if sentence .0 Less than 10,if The conditional expression of the statement does not hold , Don't execute break;
 perform y=y+x; y=0+1=1
 Turn to the expression 3, perform x++,  x=x+1=1+1=2. The loop condition x<=10 establish , In the first 2 Secondary cycle 

 The first 2 Secondary cycle 
	 perform if sentence .1 Less than 10,if The conditional expression of the statement does not hold , Don't execute break;
 perform y=y+x; y=1+2=3
 Turn to the expression 3, perform x++,  x=x+1=2+1=3. The loop condition x<=10 establish , In the first 3 Secondary cycle 

 The first 3 Secondary cycle 
	 perform if sentence .3 Less than 10,if The conditional expression of the statement does not hold , Don't execute break;
 perform y=y+x; y=3+3=6
 Turn to the expression 3, perform x++,  x=x+1=3+1=4. The loop condition x<=10 establish , In the first 4 Secondary cycle 

 The first 4 Secondary cycle 
	 perform if sentence .6 Less than 10,if The conditional expression of the statement does not hold , Don't execute break;
 perform y=y+x; y=6+4=10
 Turn to the expression 3, perform x++,  x=x+1=4+1=5. The loop condition x<=10 establish , In the first 5 Secondary cycle 

 The first 5 Secondary cycle 
	 perform if sentence .10 be equal to 10,if The conditional expression of the statement holds , perform break, Out of the loop .
 from break Jump to for The next statement of a statement . perform printf(“%d   %d”,y,x);
 Output the current y And x. The result is 10   5


14.
#include<stdio.h>
 main( )
 {
      char ch;
    ch=getchar( ); 
    switch(ch)
      {
      case  ‘A’ : printf(“%c”,’A’);
         case  ‘B’ : printf(“%c”,’B’); break;
         default: printf(“%s\n”,”other”);
 } }
 When entering letters from the keyboard A when , The running result is ：AB
 See the textbook for details p103,switch sentence 
switch（ expression ）
{
      	case   Constant 1 ： sentence 1 
    case   Constant 2 ： sentence 2 
       ┇    ┇       ┇
    case   Constant n ： sentence n 
    default     :   sentence n+1 
} 
 Where the expression , Constant 1,…, Constant n Both are integer or character types 

case It is equivalent to giving the entry and starting position of the execution program , If you find a matching constant , Then start from here to execute the program , No longer match constants , Until I met break or switch end 

 The process of this problem ：
 First receive a character from the keyboard ’A’ And put it in the variable ch in .
 perform switch sentence .Switch The following conditional expression is ch, Therefore, the value of the expression is the character ’A’.  Use characters ’A’ In turn with the following  case Constant match in .
 With the first 1 individual case The constant after matching , Then execute the program from the following statement （ No more matching during execution .） So first execute printf(“%c”,’A’), Output... On the screen A; Go on printf(“%c”,’B’), Output... On the screen B; And go on with it break,  Jump out at this time switch sentence .


15.  
#include <stdio.h>
main( )
{
     int a=1,b=0;
  scanf(“%d”,&a);
switch(a)
{
      case 1: b=1;break;
case 2: b=2;break;
default : b=10;}
printf("%d ", b);
}
 If keyboard input 5, The running result is ：10
 The process of this problem ：
 First use scanf The function is a variable a The assignment is 5.
 perform switch sentence .switch The following conditional expression is a, So the value of the expression is 5.  use 5 In turn with the following  case Constant match in . No matching constants found , So two case The following statements are not executed . perform default Subsequent statements b=10; take 10 Assigned to a variable b.
 Output variables b, The result is 10



16. 
#include <stdio.h>
 main()_
{
     char grade=’C’;
  switch(grade)
        {
     
case ‘A’: printf(“90-100\n”);
          case ‘B’: printf(“80-90\n”);
case ‘C’: printf(“70-80\n”);
case ‘D’: printf(“60-70\n”); break;
case ‘E’: printf(“<60\n”); 
default : printf(“error!\n”);
           }   
 }
 The running result is ：
70-80
60-70
 The process of this problem ：
 First receive a character from the keyboard ’C’ And put it in the variable grade in .
 perform switch sentence .switch The following conditional expression is grade, Therefore, the value of the expression is the character ’C’.  Use characters ’C’ In turn with the following  case Constant match in .
 With the first 3 individual case The constant after matching , Then execute the program from the following statement （ No more matching during execution .） So first execute printf(“70-80\n”);, Output... On the screen 70-80, And line breaks ; Go on printf(“60-70\n”), Output... On the screen 60-70, And line breaks ; And go on with it break,  Jump out at this time switch sentence .


17.
#include <stdio.h>
main()
{
     int y=9;
 for(;y>0;y- -)
if(y%3==0) 
{
     printf(%d”,- -y); 
}
}
 The running result is ：
852

 See the textbook for details p53, Self increasing and self decreasing symbol 
 This question   expression 1 Omitted , expression 2（ The loop condition ） by y>0, expression 3 by y--
 initial value y by 9, The loop condition （ That's the expression 2）y>0 establish , Into the circulatory body 

 The first 1 Secondary cycle 
	 perform if sentence .9%3==0,if The conditional expression of the statement holds , perform printf(%d”,- -y), namely y First self reduction 1 Turn into 8, And then in the output , So output on the screen 8
 Turn to the expression 3, perform y--,  y=y-1=8-1=7. The loop condition y>0 establish , In the first 2 Secondary cycle 

 The first 2 Secondary cycle 
	 perform if sentence .7%3 Not for 0,if The conditional expression of the statement does not hold , Don't execute printf(%d”,- -y)
 Turn to the expression 3, perform y--,  y=y-1=7-1=6. The loop condition y>0 establish , In the first 3 Secondary cycle 

 The first 3 Secondary cycle 
	 perform if sentence .6%3==0,if The conditional expression of the statement holds , perform printf(%d”,- -y), namely y First self reduction 1 Turn into 5, And then in the output , So output on the screen 5
 Turn to the expression 3, perform y--,  y=y-1=5-1=4. The loop condition y>0 establish , In the first 4 Secondary cycle 

 The first 4 Secondary cycle 
	 perform if sentence .4%3 Not for 0,if The conditional expression of the statement does not hold , Don't execute printf(%d”,- -y)
 Turn to the expression 3, perform y--,  y=4-1=3. The loop condition y>0 establish , In the first 5 Secondary cycle 

 The first 5 Secondary cycle 
	 perform if sentence .3%3==0,if The conditional expression of the statement holds , perform printf(%d”,- -y), namely y First self reduction 1 Turn into 2, And then in the output , So output on the screen 2
 Turn to the expression 3, perform y--,  y=y-1=2-1=1. The loop condition y>0 establish , In the first 5 Secondary cycle 
 The first 6 Secondary cycle 
	 perform if sentence .1%3 Not for 0,if The conditional expression of the statement does not hold , Don't execute printf(%d”,- -y)
 Turn to the expression 3, perform y--,  y=1-1=0. The loop condition y>0 Don't set up , The loop ends .


18. 
#include <stdio.h>
main()
{
     int i,sum=0;  i=1;
do{
     sum=sum+i; i++; }while(i<=10);
printf(“%d”,sum);
}
 The running result is ： 55




19.
#include <stdio.h>
#define N 4
main()
{
     int i;
 int x1=1,x2=2;
printf("\n");
for(i=1;i<=N;i++)
{
     printf("%4d%4d",x1,x2);
if(i%2==0) 
printf("\n");
x1=x1+x2;
x2=x2+x1;
}
}
 The running result is ：
1   2   3   5
8   13  21  34

 This question   First, assign an initial value to an integer variable x1=1,x2=2
 expression 1 by i=1, expression 2（ The loop condition ） by i<=N namely i<=4, expression 3 by i++
 Initial value of cyclic variable i by 1, The loop condition （ That's the expression 2）i<=4 establish , In the first 1 Secondary cycle 

 The first 1 Secondary cycle 
	 perform printf("%4d%4d",x1,x2); So output on the screen 1    2
 perform if sentence .1%2 Not for 0,if The conditional expression of the statement does not hold , Don't execute printf("\n");
 perform x1=x1+x2=1+2=3; here x1 The value in has changed to 3
 perform x2=x2+x1=2+3=5.
 Turn to the expression 3, perform i++,  i by 2. The loop condition i<=4 establish , In the first 2 Secondary cycle 
 The first 2 Secondary cycle 
	 perform printf("%4d%4d",x1,x2); So output on the screen 3    5
 perform if sentence .2%2==0,if The conditional expression of the statement holds , perform printf("\n"); Line break 
 perform x1=x1+x2=3+5=8; here x1 The value in has changed to 8
 perform x2=x2+x1=5+8=13.
 Turn to the expression 3, perform i++,  i by 3. The loop condition i<=4 establish , In the first 3 Secondary cycle 
 The first 3 Secondary cycle 
	 perform printf("%4d%4d",x1,x2); So output on the screen 8    13
 perform if sentence .3%2 Not for 0,if The conditional expression of the statement does not hold , Don't execute printf("\n");
 perform x1=x1+x2=8+13=21; here x1 The value in has changed to 21
 perform x2=x2+x1=21+13=34.
 Turn to the expression 3, perform i++,  i by 4. The loop condition i<=4 establish , In the first 4 Secondary cycle 
 The first 2 Secondary cycle 
	 perform printf("%4d%4d",x1,x2); So output on the screen 21    34
 perform if sentence .4%2==0,if The conditional expression of the statement holds , perform printf("\n"); Line break 
 perform x1=x1+x2=21+34=55; here x1 The value in has changed to 55
 perform x2=x2+x1=34+55=89.
 Turn to the expression 3, perform i++,  i by 5. The loop condition i<=4 Don't set up , End of cycle 


20
#include <stdio.h>
   main( )
   {
      int  x, y;
      for(x=30, y=0; x>=10, y<10; x--, y++)
         x/=2, y+=2;
      printf(“x=%d,y=%d\n”,x,y);
}
 The running result is ：
x=0,y=12

21. 
#include <stdio.h>
#define N 4
main( )
{
     int  i,j;
 for(i=1;i<=N;i++)
  {
     for(j=1;j<i;j++) 
	    printf(" ");
  printf("*");
  printf("\n");
}}
 The running result is ：
*
  *
*
   *

 See the textbook for details P41 Symbolic constant 
 Define symbolic constants with macro processing instructions N by 4, During the compilation process , encounter N It is regarded as an integer 4.

 Outer layer for loop , expression 1 by i=1, expression 2（ The loop condition ） by i<=N, expression 3 by  i++
 Inner layer for loop , expression 1 by j=1, expression 2（ The loop condition ） by j<i,    expression 3 by j++

 First, calculate the expression of the outer loop 1,i by 1, Make the cycle condition i<=4 establish , Into the outer layer for The loop body 

 Outer layer for Circle the first 1 Time      here i by 1
	 Inner circulation j=1, Make the cycle condition j<i Don't set up , Therefore, the inner loop body... Is not executed （ Don't output spaces ）
	 perform printf("*");
	 perform printf("\n"); Line break 
	 At this point, the outer loop body executes , The expression for calculating the outer loop 3,i++, here i by 2.  Make the cycle condition i<=4
 establish , Enter the outer layer again for The loop body 
	
 Outer layer for Circle the first 2 Time     here i by 2
	 Inner circulation j=1, Make the cycle condition j<i establish 
 The first 1 Execute inner loop body once printf(" ");
 Execute inner loop expression 3,j++ by 2,j<i Don't set up , Out of the inner circle 

	 perform printf("*");
	 perform printf("\n"); Line break 
	 At this point, the outer loop body executes , The expression for calculating the outer loop 3,i++, here i by 3.  Make the cycle condition i<=4
 establish , Into the outer layer for The loop body 

 Outer layer for Circle the first 3 Time     here i by 3
	 Inner circulation j=1, Make the cycle condition j<i establish 
 The first 1 Execute inner loop body once printf(" ");
 Execute inner loop expression 3,j++ by 2,j<i establish , Execute the inner loop again 
 The first 2 Execute inner loop body once printf(" ");
 Execute inner loop expression 3,j++ by 3,j<i Don't set up , Out of the inner circle 
	 perform printf("*");
	 perform printf("\n"); Line break 
	 At this point, the outer loop body executes , The expression for calculating the outer loop 3,i++, here i by 4.  Make the cycle condition i<=4
 establish , Into the outer layer for The loop body 

 Outer layer for Circle the first 4 Time     here i by 4
	 Inner circulation j=1, Make the cycle condition j<i establish 
 The first 1 Execute inner loop body once printf(" ");
 Execute inner loop expression 3,j++ by 2,j<i establish , Execute the inner loop again 
 The first 2 Execute inner loop body once printf(" ");
 Execute inner loop expression 3,j++ by 3,j<i establish , Execute the inner loop again 
 The first 3 Execute inner loop body once printf(" ");
 Execute inner loop expression 3,j++ by 4,j<i Don't set up , Out of the inner circle 
	 perform printf("*");
	 perform printf("\n"); Line break 
	 At this point, the outer loop body executes , The expression for calculating the outer loop 3,i++, here i by 5.  Make the cycle condition i<=4
 Don't set up , Jump out of the outer layer for The loop body