10424 - Love Calculator

[Wei-Ming Chen]

I got WA on problem 10424
Here are my method.

1. If scan EOF return 0
2. If line one = line two = 0 print a blank line
3. If line one = 0 or line two = 0 print 0.00%
4. If line one <= line two print 100.00%
5. Else print line two / line one

Can some tell me where is wrong?

[Eddie]

I can't solve a simple portion of 10424.........the last portion of the problem tells that to add a no. like this---------if the no. is 999 then 9+9+9=27.again 2+7=9.then stop CAN ANY ONE HELP ME......................

[Eddie]

I can't solve a simple portion of 10424.........the last portion of the problem tells that to add a no. like this---------if the no. is 999 then 9+9+9=27.again 2+7=9.then stop how could I solve it...CAN ANY ONE HELP ME......................

[Donotalo]

my output code:

Code: Select all

if (!n1 || !n2)
	cout << endl;
else if (n1 < n2)
	printf("%.2f%%\n", 100*double(n1)/n2);
else
	printf("%.2f%%\n", 100*double(n2)/n1);

i'm getting PE, why?

[Darko]

You are missing a space character. But I am not sure why this works - maybe there are no cases with n1 or n2 being zero.

[Donotalo]

yes, u r right. there is a space character missing.

but i dont understand what u didnt understand! if any n is 0, i think there shud be a blank line?

i tried with 0.00% (not 0.00 %) too and got PE. probably both works.

[Darko]

In my solution I print "100.00 %" for both being 0. And "0.00 %" if one of them is zero. So there are probably no cases where either of them is zero.

[Tahasin]

#include<iostream>
using namespace std;
#include<stdio.h>
#include<string.h>
int sum2(int p)
{
int q,s=0;
do
{
q=p%10;s+=q;p/=10;
}while(p!=0);
return s;
};
int sum1(char a[100])
{
int i,m,l,n=0;
for(i=0;a!='\0';i++)
{
m=a-'a'+1;
if(m>0)l=m;
else l=m+32;
if(l>=1 && l<=26)n+=l;
}
return n;
};
main()
{
char k[100],t[100];
int x,y,z;
float e,f,g;
while(cin>>k>>t)
{
x=sum1(k);
y=sum1(t);
do{x=sum2(x);}while(x>9);
do{y=sum2(y);}while(y>9);
if(x>y){e=x;f=y;}
else {e=y;f=x;}
g=(f/e)*100;
printf("%.2f %%\n",g);
}
return 0;
}

[Iffat]

Code: Select all

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<ctype.h>
int findnum(char str[])
{
		long i,j,l;
		int s;
		l=strlen(str);
		i=0;s=0;
		while(i<l)
		{	
			
			if(str[i] > 96 && str[i] <123)
			{
				j = str[i]-96;
				s = s + j ;
				i++;
			}
			else if(str[i] > 64 && str[i] < 91)
			{
				j= str[i] -64 ;
				s = s + j;
				i++;
			}
		
		 	
			else if(str[i]=='\0'){i++;}
		}
		return s;
}
int main()
{
	char str1[30],str2[30],str3[30],str4[30];
	int tt,ttt,l1,l2;
	int t,t1,t2,t3,ss,sss;
	while(gets(str1))
	{
	gets(str2);
	t1=	findnum(str1);
	t=t1;
	while(t>9)
	{   
		sprintf(str3,"%d",t);
		l1=strlen(str3);
		ss=0;
			for(tt=0;tt<l1;tt++)
			{
				ss=((str3[tt]-'0')+ss);
			}
			t=ss;
	}
	t2=findnum(str2);
	t3=t2;
	while(t3>9)
	{   
		sss=0;
		sprintf(str4,"%d",t3);
		l2=strlen(str4);
			for(ttt=0;ttt<l2;ttt++)
			{
				sss=((str4[ttt]-'0')+sss);
			}
			t3=sss;
	}

	double ratio;
	
	if(isalpha(str1[0])!=0&&isalpha(str2[0])!=0)	
	   {	 
		if(t3>=t)
			{  
			 ratio=((double)t/t3)*100;
				printf("%.2lf %c\n",ratio,'%');  
			}
		else if(t3<t)
			{	
			 ratio=((double)t3/t)*100;
			printf("%.2lf %c\n",ratio,'%');  
			}
		else if(t==0||t3==0)printf("\n");	 
	   }
	   else
	   {
	   		printf("\n");	 
	   }
	 
	 
	}
	return 0;
}

i can't find wats wrong in my code...i got TLE...plzzzz help me

[ishtiaq ahmed]

visit
http://acm.uva.es/board/viewtopic.php?t ... dbb19fef80

[daveon]

printf("%.2f % \n",res);

This line looks abit fishy to me. Try

Code: Select all

printf("%.2f %s\n",res,"%"); 

[newton]

Have you checked those forum before posting your problem?


http://online-judge.uva.es/board/viewtopic.php?t=7600,
http://online-judge.uva.es/board/viewtopic.php?t=2497 and
http://online-judge.uva.es/board/viewtopic.php?t=2166)


always tray to seek your solution first if dont get then post.

[daveon]

Also, this post should go in Volume CIV.

[little joey]

Also, this post should go in Volume CIV.

Done.

To print a percentage sign with printf you can also use a doubled one in the format string:

Code: Select all

printf("%.2f %%\n",res);

[daveon]

Ah, so the % is the escape character. Cool! Thanks little Joey, your code is shorter.