Your code can be compiled, and the input output, after the first set of test data and the output of the correct answer is the same,
But all four groups of test data is wrong answer,
Hope to have a great god can help you see, the grateful orz
The first set of test data:
2
10011



Topics are as follows:
Title description
If a number (the first is not zero) read from right to left and from left to right are the same, we call it the palindrome,

For example: given a decimal number 56, add 56 to 65 (that is, the 56 read from right to left), is a palindrome, you get 121

Such as: for the decimal number 87:

STEP1:87 + 78=165
STEP2:165 + 561=726
STEP3:726 + 627=1353
STEP4:1353 + 3531=4884
Step here refers to an N in addition, the patients use the four steps to get at least 4884, palindrome

Write a program that, given a N (2 N 10 or less or less or N=16) hexadecimal number M within (100), at least after a few steps can get a palindrome, if within 30 steps (including step 30) may not get palindrome, output is Impossible! ,

Input format
Two lines, N, M,

The output format
If can get palindrome within 30 steps, output formats such as STEP=ans, the ans steps to at least get a palindrome,

Otherwise the output Impossible! ,


the building code is as follows:

 # include 
Int judge (int * p, int bit); 
Int main () 
{
Int n; 
The scanf (" % d \ n ", & amp; N); 
char ch; 
Int I=0, j, carry=0;//counter I, j and remainder carry 
Int a [200], [200] b; 
While ((ch=getchar ())! ) 
='\ n'{
If (ch>='0' & amp; & ChElse a [i++]=ch - 'a' + 10; 
}//cycle after a [I - 1) is the final number 
For (j=0; J<=I - 1; + + j) 
B [j]=[] I - j - 1;//to reverse a 
If (judge (a, I))//if the input palindrome, then output the number of 0. 
Printf (" STEP=0 \ n "); 
The else 
{
Int c [200], bit=I, ans=0;//I, but, pay attention to the array first is [0]; 
While (1) 
{
For (j=0;; J++) 
{
C [j]=(a + b [j] [j] + carry) % n; 
If (a + b [j] [j] + carry>=n)//determine whether have carry 
Carry=1; 
The else 
Carry=0; 
If (j==bit - 1)//if the last 
{
If (carry==1) 
{
Bit +=1;//digits plus 1; 
C [bit - 1]=1;//first into 1 
Carry=0;//remainder reset 
} 
break; 
} 

} 
+ + ans;//the number 1 
If (judge (c, a bit))//judge whether palindrome 
{
Printf (" STEP=% d \ n ", ans); 
return 0; 
} 
If (ans> 30)//if the number is greater than 30 
{
Printf (" Impossible! \n"); 
return 0; 
} 
For (j=0; J<=bit - 1; J++)//redefine ABC 
{
A [j]=c [j]; 
B] [bit - j - 1=c [j]; 
} 

} 


} 

return 0; 
} 
Int judge (int * p, int bit)//judge whether palindrome 
{
int i; 
for(i=0; I<=bit - 1; I++) 
{
If (p [I]!=p -i] [bit - 1) 
return 0; 
} 
return 1; 
} 

CodePudding user response:

C [j]=(a + b [j] [j] + carry) % n;

If the input is the hex you're wrong
A [j] is' a '
You calculate (' a '+' a ') % 16?

CodePudding user response:

More than a decimal your algorithm is wrong

CodePudding user response:

Oh, I see wrong
If (ch>='0' & amp; & ChElse a [i++]=ch - 'a' + 10;
Have to do special treatment

CodePudding user response:

There are holes, 44 when ans=31 still output, is amended as:

 if (judge (c, bit) & amp; & Ans<=30)//judge whether palindrome 

CodePudding user response:

I see, is likely to be the output STEP=31