Questions do something like this:
You are asked to write a program to bear on the given symbol print into the shape of an hourglass, given 17 "*", for example, requirements in the following format to print

* * * * *
* * *
*
* * *
* * * * *
"The hourglass shape", refers to the output an odd number of symbols per line; The center line all symbols; Adjacent two rows symbol number 2; Diminishing the number symbols from big to small order at first to 1, since the childhood again increasing order. The end is the same symbol.

Given any N symbols, can not form an hourglass, just asked to print out of the hourglass can drop as many symbols,

Input format:
Type in a line is given a positive integer N (1000) or less and a symbol, separated by Spaces in the middle,

The output format:
Print out the first largest hourglass shape consisting of a given symbol, the final output useless left off in a row the number of symbols,

Input the sample:
19 *
The output sample:
* * * * *
* * *
*
* * *
* * * * *
2
And my code is

 # include "stdio.h" 
Int main () 
{
Int n, I, u=0, j, sum=1, the count=0, len. 
Char c; 
The scanf (" % d % c ", & amp; N, & amp; C); 
If (n==1) 
{
Printf (" * \ n0 "); 
Goto the EX. 
} 
If (n==2) 
{
Printf (" * \ n1 "); 
Goto the EX. 
} 
If (n> 1000 | | n<=0) return 0; 
For (I=3; Sum<=n;) 
{
Sum=sum + 2 * I; 
I=I + 2; 
count++; 
} 
J=n - (sum - 2 * (I - 2)); 
Len=I - 4; 
for(i=0; i{
for(n=0; Nfor(n=0; Nprintf("\n"); 
Len=len - 2; U++; 
} 
Len=len + 4; U=u - 2; 
for(i=1; i{
for(n=0; Nfor(n=0; Nprintf("\n"); 
Len=len + 2; U -; 
} 
Printf (" % d ", j); 
EX: 
return 0; 
} 

Out why the answer is partially correct ah, for help

CodePudding user response:

For your reference

 void test (int n, char * c) 
{
Int x=n/2; 
Int num=SQRT (x); 
Int I=0, j=0, k=0; 

//part 1 
For (I=num; I> 0; I -) 
{
//space 
for(k=0; K{
Printf (" "); 
} 
For (j=2 * I - 1; J> 0; J -) 
{
Printf (" % c ", c); 
} 
printf("\n"); 
} 
//part 2 
for(i=1; i{
For (k=I; K{
Printf (" "); 
} 
for(j=0; J<2 * I + 1; J++) 
{
Printf (" % c ", c); 
} 
printf("\n"); 
} 


} 

This call
The test (99, '*');

CodePudding user response:

Reference:

 # include "stdio.h" 
Int main () 
{
Int n, I, j, sum=0, count=1;//, len; , u=0 
Char c; 
The scanf (" % d % c ", & amp; N, & amp; C); 
/* if (n==1) 
{
Printf (" * \ n0 "); 
Goto the EX. 
} 
If (n==2) 
{
Printf (" * \ n1 "); 
Goto the EX. 
} */
If (n> 1000 | | n<=0) return 0; 
For (I=3; Sum<=n;) 
{
Sum=sum + 2 * I; 
I=I + 2; 
count++; 
} 
The count=count - 1; 
//j=n - (sum - 2 * (I - 2)); 
//len=I - 4; 
For (I=count; I>=1; I -)//upper 
{
for(j=0; Jfor(j=0; J<2 * I - 1; J++) printf (" % c ", c); 
printf("\n"); 
//len=len - 2; U++; 
} 
//len=len + 4; U=u - 2; 
For (I=I + 2; i<=count; 
 lower i++)//{
for(j=0; Jfor(j=0; J<2 * I - 1; J++) printf (" % c ", c); 
printf("\n"); 
//len=len + 2; U -; 
} 
Printf (" % d ", n - (2 * count * count - 1)); 
//printf (" % d ", j); 
//the EX: 

return 0; 
} 


//35 * 
//* * * * * * * 
//* * * * * 
//* * * 
//* 
//* * * 
//* * * * * 
//* * * * * * * 
//4 please press any key to continue... 

CodePudding user response:

I go to a test run on the PTA, or like the original, part of the answer wrong, seems to be some kind of special case is not applicable

CodePudding user response:

Finally a printf didn't bring \ n?

CodePudding user response:

Only consider the format and special circumstances,
One can try finally have a newline,
Two can have a try if it is zero is not output,