10424 - Love Calculator

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RustB
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Posts: 16
Joined: Mon Jun 14, 2004 5:08 pm

10424 Love Calculator WA

Post by RustB »

Code: Select all

#include <stdio.h>

int firstsum(char *inp)
{
	int i=0,sum=0;
	while(inp[i])
	{
	     if(inp[i]>='a' && inp[i] <='z')
	     	sum+=inp[i]-'a'+1;
	     else if(inp[i]>='A' && inp[i] <='Z')
	     	sum+=inp[i]-'A'+1;
	     i+=1;
	}
	return sum;
}
int sumnum(int num)
{
	int sum=0;
	while(num)
	{
	     sum+=num%10;
	     num/=10;
	}
	return sum;
}

int main()
{
	char x[30],y[30];
	int sumx,sumy;
	float ratio;
	while(!feof(stdin))
	{
          gets(x);
	     sumx=firstsum(x);
	     while(sumx>=10)
	          sumx=sumnum(sumx);
	     gets(y);
	     sumy=firstsum(y);
	     while(sumy>=10)
	          sumy=sumnum(sumy);
	     if(sumx > 0 && sumy > 0)
	     {
     		if(sumx>sumy)
     	     	ratio = ((float)sumy/(float)sumx)*100.0f;
     	     else
     	     	ratio = ((float)sumx/(float)sumy)*100.0f;
     	}
     	else if(sumx==0 && sumy == 0)
     	{
			printf("\n");
			continue;
		}
		else
		{
			printf("0.00 %%\n");
			continue;
		}
		printf("%.2f %%\n",ratio);
	}
	return 0;
}
I have checked it with every input I can find and it gives the right answer.
-I print blank line if both sums are 0
-I print upto 2 decimal points accuracy

Are there any special inputs?

sumankar
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Post by sumankar »

Hi,

What happens when one of the sums is non-zero and the other is zero?
One more thing, I think that sum of digits as you have implemented can be
simplified by taking it modulo-9.

Code: Select all

31 = 3 + 1 = 4
31 % 9 = 4
Regards,
Suman.

RustB
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Posts: 16
Joined: Mon Jun 14, 2004 5:08 pm

Post by RustB »

sumankar wrote:Hi,

What happens when one of the sums is non-zero and the other is zero?
The output in this case is 0.00 %. Is that correct?

sumankar
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Location: calcutta
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Post by sumankar »

Hi,

Yeah!

But i dont think your code handles the case where both inputs are empty
correctly.Check that out.

Regards,
Suman.

RustB
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Posts: 16
Joined: Mon Jun 14, 2004 5:08 pm

Post by RustB »

If either or both inputs are empty it just prints a blank line.

Can you give me the specific inputs for which you think this program is generating incorrect outputs?

sumankar
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Location: calcutta
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Post by sumankar »

Hi,

No I am clueless.Solved this one long long back.Forgot most tricks ...if there was one.

But i'll look into it.

Regards,
Suman.

sumankar
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Location: calcutta
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Post by sumankar »

One final try:

what is the first is a valid string and next input is EOF?
how do tackle that?

Regards,
Suman.

RustB
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Posts: 16
Joined: Mon Jun 14, 2004 5:08 pm

Post by RustB »

My original code would exit if either input was EOF.

I changed it such that even if the second line is EOF, it continues calculation with the second input as a null string, still WA.

Thanks for all the help. I think this is not a problem with my logic, but an input validation issue. I do not have the patience to sit and debug this either. I will not try too much, it is a waste of time.

Ryan Pai
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Joined: Fri Jul 04, 2003 9:59 am
Location: USA
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Post by Ryan Pai »

After running your code on a few different inputs, you didn't get any answer different than me, but yours did print out an extra line of input. You should check to make sure that at least two lines exist.

It is a very bad habbit to check for end of file in programming contests, many beginners fail due to mistakes like this, especially since sometimes there is an end of line after the last data set, and sometimes there isn't.

So it's usually a better idea to try to read a dataset past where the last one ends, and if you can't read it, then bail. For example:

Code: Select all

while(gets(x) && gets(y)){
  //...
}
I'm always willing to help, if you do the same.

Sedefcho
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Location: Bulgaria

Post by Sedefcho »

Hi, RustB !

Your program is almost working.

I have just made some minor changes to the first 10-15 lines of
your main() function. Just repeat them and you'll get Aceepted.

One of the problems fixed was this one. You have 2*N lines
and the last line ends with EOL. Then an EOF follows.
In that case your program was printing an additional
output line containing 0.00%. Which resulted in WA of course.

Here is your code after my changes.

Code: Select all

int main() 
{ 
   char x[30],y[30]; 
   int sumx,sumy; 
   float ratio; 
   // while(!feof(stdin))
   while(gets(x))
   {
   gets(y);
   if ( strlen(x)==0 ) break; 
        // gets(x); 
        sumx=firstsum(x); 
        while(sumx>=10) 
             sumx=sumnum(sumx); 
        // gets(y); 
        sumy=firstsum(y); 
        while(sumy>=10) 
             sumy=sumnum(sumy);
        
        // REST OF main() REMAINS UNCHANGED !  

Good luck !

jaracz
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Posts: 79
Joined: Sun Sep 05, 2004 3:54 pm
Location: Poland

Post by jaracz »

I solve this problem without any suggestions which took place above.
my loop looks like this:

Code: Select all

while(gets(name1) && gets(name2))
    {
        ...(calculate both values)
        if(value1>value2)ratio = value2/value1;
        else ratio = value1/value2;
        printf("%.2lf %%\n",ratio*100);
    }
you guys don't have to check if strlen(a)==0 ... or any value == 0 and so on...

You'd better check your alphabet before post;)
Remember that ..opqrstuvxyz
Hope it helps!!
keep it real!

kakashi
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Posts: 5
Joined: Thu May 26, 2005 10:58 am

10424 Love Calculator

Post by kakashi »

Code: Select all

#include<iostream>
#include<cstdio> 
#include<string>
#include<cctype>

using namespace std;
int main(){
    string name1,name2;
    int length1,length2;
    int i;
    int num1,num2;         
    double ans1,ans2; 
    
    for(cin >> name1 >> name2;cin;cin >> name1 >> name2){
            num1=0;
            num2=0; 
            length1=name1.length();
            length2=name2.length();
            
            for(i=0;i<length1;i++){
                if(islower(name1[i]))    
                   name1[i]=(name1[i]-32);
                                      
                num1+=int(name1[i])-64;
            }
            for(i=0;i<length2;i++){
                if(islower(name2[i]))       
                   name2[i]=(name2[i]-32);
                                    
                num2+=int(name2[i])-64;
            }
            
            
            while(num1>9||num2>9){ 
                ans1=0; 
                while(num1){
                      ans1+=num1%10;
                      num1/=10;
                }
                num1=int(ans1); 
            
                ans2=0; 
                while(num2){
                      ans2+=num2%10;
                      num2/=10;
                }
                num2=int(ans2); 
            } 
            printf("%.2f %%\n",(ans1>ans2 ? (ans2/ans1)*100 : (ans1/ans2)*100)); 
    }
}
I try many input data...
but I can`t find which is wrong .
plz help me :cry:

Niaz
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Posts: 77
Joined: Fri Dec 17, 2004 11:06 am
Location: East West University, Dhaka, Bangladesh
Contact:

Post by Niaz »

You have done a very simple mistake. Just change your input taking methods, i.e. consider spaces. I hope you got my point !

Thanks for trying my problem. :D
Please join The ACM Solver Group at Yahoo
http://groups.yahoo.com/group/acm_solver/

kakashi
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Posts: 5
Joined: Thu May 26, 2005 10:58 am

Post by kakashi »

thanks......
I try to change input taking methods......
using getline and it does work... :D

ashikzinnatkhan
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Posts: 8
Joined: Wed Jan 25, 2006 6:25 pm
Location: Dhaka, Bangladesh

10424 love calculator , WA , please help me

Post by ashikzinnatkhan »

Here is my code:




#include <stdio.h>


char name1[30] , name2[30] ;

void main()
{

char letter[26] = { 'a' , 'b' , 'c' , 'd' , 'e' , 'f' , 'g' , 'h' , 'i' , 'j' , 'k' , 'l' , 'm' , 'n' , 'o' , 'p' ,
'q' , 'r' , 's' , 't' , 'u' , 'v' , 'w' , 'x' , 'y' , 'z' };

int sum1 , sum2 , i , j , temp[3] , temp2;

float ratio , one , two;

char ch;


while( (scanf("%s",name1)) == 1 )
{
scanf("%s",name2);

sum1 = 0;
sum2 = 0;

for(i=0 ; name1 ; i++)
{
if(name1>=65 && name1<=90) ch = letter[(int) name1 - 65];
else ch = name1;

for(j=0; j<26 ; j++)
{
if(ch == letter[j]) sum1 = sum1+j+1;
}

}

for(i=0 ; name2 ; i++)
{
if(name2>=65 && name2<=90) ch = letter[(int) name2 - 65];
else ch = name2;

for(j=0; j<26 ; j++)
{
if(ch == letter[j]) sum2 = sum2+j+1;
}

}


if( sum1 > 10 )
{
while(sum1 > 10)
{

temp2 = sum1;

temp[0] = temp2%10;

temp2 = temp2 - temp[0];
temp2 = temp2/10;
temp[1] = temp2%10;

temp2 = temp2 - temp[1];
temp[2] = temp2/10;

sum1 = temp[0] + temp[1] + temp[2] ;

}

}



if( sum2 > 10 )
{
while(sum2 > 10)
{

temp2 = sum2;

temp[0] = temp2%10;

temp2 = temp2 - temp[0];
temp2 = temp2/10;
temp[1] = temp2%10;

temp2 = temp2 - temp[1];
temp[2] = temp2/10;

sum2 = temp[0] + temp[1] + temp[2] ;

}

}


one = (float) sum1;
two = (float) sum2;

if(one > two) ratio = two/one ;
else ratio = one/two ;

printf("%.2f %%\n", ratio*100.0);


}

}







It is giving WA.
Please help me.
Ashik

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