10931 - Parity

[scidong]

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#include<iostream.h>
#include<string.h>
char a[30];
void dtob(int c){
	int i=0;
	while(c){
		a[i]=c%2;
		c=c/2;
		i++;
	}
}

void main(){
	int c,i,p,cn;
	while(cin >> c){
		if(c==0) break;
		if(c==1) cout << "The parity of 1 is 1 (mod 2)." << endl;
		else{
			dtob(c);
			cout << "The parity of ";
			cn=0;
			for(i = 0; i<30; i++){
				if(a[i] == 1){
					p=i;
					cn++;
				}
			}for(i = p; i>=0; i--) cout << int(a[i]);
			cout << " is " << cn << " (mod 2)." << endl;
		}
	}
}

plz help me...

[jjtse]

hey scidong,

your posts was a long time ago. I recently solved that problem. If you still need help on that problem, post a reply and I'll help you out. Otherwise I'll assume you already figured it out.

[scidong]

Uh... plz help me sir.

[jjtse]

This is a simple problem actually. Since


1 ≤ I ≤ 2147483647


integers will not be big enough to hold that.
Take a look at this: http://home.att.net/~jackklein/c/inttypes.html#limits

I guess for a quick fix, just replace all your "int" types with "long" or "unsigned long" Either of those will work.

[ayon]

int is enough for this problem, and for uva judge there is NO difference between int and long, my ac program everywhere used int data type.

[Martin Macko]

plz help me...

Try the following input:

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2147483647
0

The correct answer is:

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The parity of 1111111111111111111111111111111 is 31 (mod 2).

However, your solution says that the parity of 2147483647 is 30.

[hosnayen]

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/*
	Author : Hosnayen Alam Siddiquee.
	University of Science & Technology Chittagong (USTC)
	CSE- 11th Batch
	E-mail: hosnayen_alam@yahoo.com
	Bangladesh	                          Date:	20/08/2010
*/

#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<algorithm>
#include<set>
#include<queue>
#include<stack>
#include<list>
#include<iostream>
#include<fstream>
#include<numeric>
#include<string>
#include<vector>
#include<cstring>
#include<map>
#include<iterator>

using namespace std;

int main()
{
 long I,count=0;

 while(scanf("%ld",&I)==1 && I){
   long res[100];
    long i=0,j=0;
	 while(I>0){
		   res[i]=I%2;
		   j++;
		   if(res[i]==1){
			count++;
		   }
        I=I/2;
      i++;
	 }

     printf("The parity of ");

	 if(res[0]==0){
          reverse(res,res+j);
	 }
	 
	 for(i=0;i<j;i++){
        printf("%ld",res[i]);
	 }
	 printf(" is %ld (mod 2).\n",count);
	 count=0;
	 
 }

return 0;
}

[helloneo]

Try this

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11

My output

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The parity of 1011 is 3 (mod 2).

Remove your code after AC

[taufique]

I just can't understand what can be the reason of WA here.
Its a simple code but WA.
anybody can help me??
Here is my code

#include <iostream>
#include <cstring>
using namespace std;

int c;

char* calculate(int n)
{
int i = 0,j;
c = 0;
char bin[50];

if(n%2) c++;
bin[i++] = (n%2)+48;
n /= 2;

while(n)
{
if(n%2) c++;
bin[i++] = (n%2)+48;
n /= 2;
}
bin = '\0';

for(i = 0,j = strlen(bin)-1 ; i < j ; i++,j-- )
{
char temp = bin;
bin = bin[j];
bin[j] = temp;
}

return bin;
}

int main()
{
int n;
char bin_n[50];
while(cin >> n)
{
if(n == 0) break;
else
{
strcpy(bin_n,calculate(n));
cout << "The parity of " << bin_n << " is " << c << " (mod 2)." << endl;
}
}
return 0;
}

[Jehad Uddin]

hello ghana,
change ur pointer calculation

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char bin[50];
void calculate(int n)
{

use code tags while posting code

[DD]

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/*
	Author : Hosnayen Alam Siddiquee.
	University of Science & Technology Chittagong (USTC)
	CSE- 11th Batch
	E-mail: hosnayen_alam@yahoo.com
	Bangladesh	                          Date:	20/08/2010
*/

#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<algorithm>
#include<set>
#include<queue>
#include<stack>
#include<list>
#include<iostream>
#include<fstream>
#include<numeric>
#include<string>
#include<vector>
#include<cstring>
#include<map>
#include<iterator>

using namespace std;

int main()
{
 long I,count=0;

 while(scanf("%ld",&I)==1 && I){
   long res[100];
    long i=0,j=0;
	 while(I>0){
		   res[i]=I%2;
		   j++;
		   if(res[i]==1){
			count++;
		   }
        I=I/2;
      i++;
	 }

     printf("The parity of ");

	 if(res[0]==0){
          reverse(res,res+j);
	 }
	 
	 for(i=0;i<j;i++){
        printf("%ld",res[i]);
	 }
	 printf(" is %ld (mod 2).\n",count);
	 count=0;
	 
 }

return 0;
}

Your code has some problems when reversing. You may try to print res in reverse order.

[NaiLuJ]

can someone help me? i don't know why i get WA for this
thanks before

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code removed , got AC thansk to brian :)

[brianfry713]

change n from long to int.

[NaiLuJ]

thanks brian, i got AC

[cse.mehedi]

Any one help me plz!!!

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AC