Printing prismatic shapes with C language. Thank you!

I've solved this problem before, now I'll figure it out for you!!

There are several ways to do this.

The simplest way to do this is to use the printf function to output it line by line. Of course, you can also use the following method.

But it's not so easy to understand.

#include

int main()

char a='*';

int i,j,k,n;

j=1;record line number;

i=1;Record the number of * signs per line.

k=8;Record the number of spaces in each row.

The n-value is used to temporarily save the k-value. Used in a for loop.

while(j<=4)

i=1+2*(j-1);

k=8-2*j;

n=k;for(;n>=0;n--)

printf(" ");

for(;i>=1;i--)

printf("* ");

printf("");

j++;while(j<=7)

i=7-2*(j-4);

k=2*(j-4);

n=k;for(;n>=0;n--)

printf(" ");

for(;i>=1;i--)

printf("* ");

printf("");

j++;return 0;

Attached is the method of using the printf function.

#include

int main()

printf(" *");

printf(" * n");

printf(" * n");

printf("* *n");

printf(" * n");

printf(" * n");

printf(" *");

return 0;

If you don't understand, please contact me.

I'll talk about the second, in fact, the first and the second ideas are the same.

1.The outside for loop, use i to control the number of rows, the input integer is n, then the number of rows of the whole graph is 2*n-1, and the number of columns is the same, you can see, that is, the number of rows and the number of columns are the same, so the inner loop and the outer loop are both 2*n-1, if you don't believe it, you can check the number of rows and columns of the graph.

is to control the lower half. Because 0<=j<=2*n-1, that is, j can be greater than n-1, or it can be smaller than the middle column of the real graph, and the whole graph is symmetrical about this column. When i = n, the principle is the same, as i increases, each row will be two fewer digits, and the position of the fewer digits will be symmetrical with respect to the middle column.

Therefore, when abs(n-1-j)<2*n-1-i, i>=n, every time i increases by 1, 2*n-1-i will decrease by 1, abs(n-1-j) will decrease by two columns, and the two columns that are reduced are about intermediate symmetry, you can calculate whether this is the truth.

Is there anything else you don't understand?

I didn't write it, but I can still read it.

You can use arithmetic to deduce the number of spaces and * in each line.

And then there's the nasty simulation.

#include

#include

#include

int main()

int i,j,k=1;

for(i=0;i<10;i++,k+=2){for(j=0;j<10-i;j++)

printf(" ");

for(j=0;j

Do you want to print a triangle, the above is to print a triangle**-Internet of Things School-Enterprise Alliance Technology Department.

The pseudo-** method determines whether it is a leap year (the big year is not involved here), because you are not sure of the language you use.

The return value is true, indicating that the year corresponding to the input parameter is a leap year. otherwise not a leap yearboolean whether it is a leap year (int year) elsereturn boo; }

#include

void main()

for(i=4;i<=7;i++) print the bottom half} You can also take it up a bit and print an empty diamond. This one of mine is simple enough, just use the for loop.

Haha, the second floor is right, a few lines of ** can be fully realized.

I don't know if it's shorter than what I wrote.

void main()}

#include

#include

int main()

for(i=1;i<=i;i++)

for(k=k;k<=2*n;k++)

printf("");

m-=1;k+=1;

i+=1;return 0;

To you the first half ... Think for yourself about the second half ... All right?

Please explain, who is going to explain to this master, is it so troublesome to write on the first floor? It would be nice to write simply.

First of all, open the C language machine in your computer and create a new source file, as shown in the figure below.

Then write **, as shown in the figure below, and the source ** is as follows: include Use C to loop to draw a diamond of numbers.

Continue to enter **, as shown in the figure below, for(i=1; i<=n-1;i++) outputs the lower half of the triangle if(i==n-1)printf("1");else{for(t=1;t<=n-i;t++)

Finally, click Compile and Run, as shown in the figure below, and then type n to see the result, diamond pattern.

Loop output, as long as the number of spaces is determined.

int main()

for (j = 0; j < 2 * i+1; j++)printf("");

for (i = 0; i < line - 1; i++)for (j = 0; j < 2 * line - 1 - i) -1; j++)

printf("");

system("pause");

return 0;

This one is a solid diamond-shaped **.

*It is easier to understand the output with the following statements, mainly the nested statements of the loop structure play a role, pay attention to the increasing relationship between spaces and asterisks* and you can output as many lines as you need. Refining! #include ""

void main()

printf(" ");

for(i=0;i<=n;i++)

#include

main()

printf("");After the * of the ith line is output, the line wrap executes the loop of the i+1 line.

The following procedure performs an inverted star triangle.

for(i=0;i<(m-1)/2;i++) Remaining number of lines m-(m+1) 2=(m-1) 2

printf("");}

* Manhattan Distance Principle solves the output diamond problem.

#include

using namespace std;

int main()

cout<}return 0;}

Tan Haoqiang's fourth edition of post-class question answers.

#include

int main()

for(b=1;b<=2*i-1;b++) controls the number of * per row, printf("");

Last 3 lines, bottom half.

for(i=n-1;i>=1;i--) control the number of rows in the first four rows for(b=1; b<=2*i-1;b++) controls the number of * per row, printf("");

printf("");

return 0;}

Extension: Print a diamond (method 1).

#include

main()

for(row=6;row<10;row++)