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Re: 10424 - Love Calculator

Posted: Sat Dec 06, 2014 3:50 pm
Input

Code: Select all

shanto
saima
Acc Output

Code: Select all

71.43 %

Re: 10424 - Love Calculator

Posted: Sat Jan 17, 2015 6:53 pm
As i know it is old thread but most of people want to know the code for creating a love calculator and i can help. I have created a web calculator in php and js located here and it is based on a new code and according to my users it is showing correct results. If you want to learn then i can help.

Re: 10424 - Love Calculator

Posted: Mon Sep 07, 2015 10:08 am
Hi,
My code isn't working for the following critical input given in uDebug:

Code: Select all

HgU
Alld
Running this input in my code, I found the output:

Code: Select all

50.00 %
I know the accepted output is 22.22 % . But is this correct? Calculating the input data manually, I found out the result should be 50.00 %. Though I might be wrong. All the other inputs are working correctly I guess. Please help.

Below is my solution in c++:

Code: Select all

#include<stdio.h>
#include<string.h>

int lc(int n)
{
if(n>=10)
{
int sum=0;
while(n)
{
sum+=(n%10);
n/=10;
}
return lc(sum);
}
}

int main()
{
int i,sum1,sum2;
float res;
char n1[30],n2[30];
while(gets(n1))
{
gets(n2);
sum1=sum2=0;
for(i=0;n1[i]!='\0';i++)
{
if(n1[i]>=97&&n1[i]<=122)
sum1+=(n1[i]-96);
else if(n1[i]>=65&&n1[i]<=90)
sum1+=(n1[i]-64);
}
for(i=0;n2[i]!='\0';i++)
{
if(n2[i]>=97&&n2[i]<=122)
sum2+=(n2[i]-96);
else if(n2[i]>=65&&n2[i]<=90)
sum2+=(n2[i]-64);
}
sum1=lc(sum1);
sum2=lc(sum2);
if(sum1<sum2)
res= (float)sum1*100/(float)sum2;
else
res= (float)sum2*100/(float)sum1;
printf("%.2f %%\n",res);
}
return 0;
}