10424 - Love Calculator

[galois_godel]

I think it's a easy problem and I got AC when I submit first. But after REJUDGEMENT. The judge tell me that my code is wrong?
Can u tell me why? thank u?
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
int chartonum(char *c)
{
int len=strlen(c);
int i;
int result=0;
for(i=0;i<len;i++)
{
if(c[i]>='a' && c[i]<='z')
result+=c[i]-'a'+1;
if(c[i]>='A' && c[i]<='Z')
result+=c[i]-'A'+1;
}
while(result>=10)
{
int temp=0;
while(result)
{
temp+=result%10;
result=result/10;
}
result=temp;
}
return result;
}


int main()
{
char a[30],b[30];
while(cin.getline(a,25))
{
double r;
//cin.getline(a,25);
cin.getline(b,25);
scanf("%s",b);
int numa=chartonum(a);
int numb=chartonum(b);
if(numa==numb)
r=1.0;
else if(numa>numb)
r=double(numb)/(double)(numa);
else
r=double(numa)/(double)(numb);
r=r*100.0;
cout.setf(ios::fixed);
cout.precision(2);
cout<<r<<" %"<<endl;
}


return 0;
}

[galois_godel]

I think it's a easy problem and I got AC when I submit first. But after REJUDGEMENT. The judge tell me that my code is wrong?
Can u tell me why? thank u?
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
int chartonum(char *c)
{
int len=strlen(c);
int i;
int result=0;
for(i=0;i<len;i++)
{
if(c[i]>='a' && c[i]<='z')
result+=c[i]-'a'+1;
if(c[i]>='A' && c[i]<='Z')
result+=c[i]-'A'+1;
}
while(result>=10)
{
int temp=0;
while(result)
{
temp+=result%10;
result=result/10;
}
result=temp;
}
return result;
}


int main()
{
char a[30],b[30];
while(cin.getline(a,25))
{
double r;
//cin.getline(a,25);
cin.getline(b,25);
scanf("%s",b);
int numa=chartonum(a);
int numb=chartonum(b);
if(numa==numb)
r=1.0;
else if(numa>numb)
r=double(numb)/(double)(numa);
else
r=double(numa)/(double)(numb);
r=r*100.0;
cout.setf(ios::fixed);
cout.precision(2);
cout<<r<<" %"<<endl;
}


return 0;
}

[Bistromath]

Hi,

This problem seems easy but i think i have excatly the same problem as u.
Is there any special case to consider, or is it a precision problem
Any tip or I/O data appreciated
Thanks

[turuthok]

galois_godel, I tried to grab your code and ran it ... It looks like something is wrong when you're reading the input (words) ...

I tried with simple input file containing:
saima
shanto

Your solution returned correct answer:
71.43 %

However, just added two more words:
saima
shanto
Pakistan
India

Your solution failed:
14.29 % ==> Looks like the love-ration between saima and Pakistan!
0.00 %

Oh yes, and one more thing that we need to be very careful, the input-description doesn't mention that each word will be in a line by itself.

-turuthok-

[Bistromath]

Oh yes, and one more thing that we need to be very careful, the input-description doesn't mention that each word will be in a line by itself.

-turuthok-

I though about that case but problem description also says that a word can contain white spaces.
how do you consider such a line :
a a b b ?

"a a" and "b b" or
"a","a" and "b","b" ?

[turuthok]

Bistromath, you are correct ... please ignore my last paragraph there ... I checked my AC code again ... it also reads one word per line ...

Sorry about that.

-turuthok-

[turuthok]

galois_godel, try to remove or comment out your 'scanf' ... it might work.

-turuthok-

[liusu]

i also meet the same proble..my code is :
[pascal]
var
table:array[0..128] of integer;
in1,in2:string;
sum1,sum2:integer;

procedure init();

var
i,x:integer;

begin
for i:=0 to 128 do
table[i]:=0;

x:=64;
for i:=1 to 26 do
table[x+i]:=i;
x:=96;

for i:=1 to 26 do
table[x+i]:=i;
end;

function len(i:integer):integer;
var
len,tem:integer;
begin
len:=1;
tem:=i;
while(tem>9) do
begin
inc(len);
tem:=tem div 10;
end;
len:=len;
end;

function getsum(s:String):integer;
var
i,tem,l:integer;

begin
tem:=0;
i:=1;
l:=length(s);

for i:=1 to l do
tem:=tem+table[ord(s[i])];
getsum:=tem;
end;


function getresult(i:integer):integer;
var
tem,x,j:integer;
begin
tem:=i;
x:=0;

while(tem>9) do
begin
if(tem>9) then
begin
for j:=0 to len(tem) do
begin
x:=x+tem mod 10;
tem:=tem div 10;
end;
tem:=x;
x:=0;
end;
end;
getresult:=tem;
end;


begin
init();
WHILE(NOT EOF) DO
begin
readln(in1);
readln(in2);
sum1:=getsum(in1);
sum1:=getresult(sum1);
sum2:=getsum(in2);
sum2:=getresult(sum2);
if(sum1>sum2) then
writeln(sum2/sum1*100:0:2,' %')
else writeln(sum1/sum2*100:0:2,' %');
end;

end.

Code: Select all

 plseae help me...thank you very much!


[Sajid]

It's so easy problem. Check your code ... debug it.
it's also easy to find out inputs and outputs for this problem. so, find some others inputs and calculate the output urself 1st.. then check with ur program...

Best of LUCK...

[Angel]

I cant find any wrong in this....... :o


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




void main()
{
int i,j,sum1=0,sum2=0;
double digit1=0.0,digit2=0.0,per;
char name1[30],name2[30];

while(gets(name1) && gets(name2))
{

for(i=0;i<strlen(name1);i++)
{
if(name1[i]>='A'&&name1[i]<='Z')
sum1+=name1[i]-'A'+1;
if(name1[i]>='a'&&name1[i]<='z')
sum1+=name1[i]-'a'+1;

}
for(i=0;i<strlen(name2);i++)
{

if(name2[i]>='A'&&name2[i]<='Z')
sum2+=name2[i]-'A'+1;
if(name2[i]>='a'&&name2[i]<='z')
sum2+=name2[i]-'a'+1;
}

if(sum1==0)digit1=0.0;
if(sum2==0)digit2=0.0;
while(sum1>0)
{
digit1+=float(sum1%10);
if(digit1>9){sum1=int(digit1);digit1=0;}
else sum1=sum1/10;

}
while(sum2>0)
{
digit2+=float(sum2%10);
if(digit2>9){sum2=int(digit2);digit2=0;}
else sum2=sum2/10;

}
if(digit1>=digit2)
{
per=(digit2/digit1)*100;
printf("%.2f %%\n",per);
}
else if(digit2>=digit1)
{
per=(digit1/digit2)*100;
printf("%.2f %%\n",per);
}


digit1=0.0;digit2=0.0;sum1=0;sum2=0;
//memset(name1,0,sizeof(name1)-1);
//memset(name2,0,sizeof(name2)-1);
}
}

[Angel]

test

[SilVer DirectXer]

[cpp]
#include <stdio.h>
#include <string.h>
struct shit{
char name[26];
int n_value;
int len_name;
};
int die1,die2;
shit n1,n2;
int i;
double ans;
void main(void)
{
while (1)
{
ans=0;
gets(n1.name);
if (feof(stdin))
break;
gets(n2.name);
if (feof(stdin))
break;
n1.len_name =strlen(n1.name );
n2.len_name =strlen(n2.name );
for (i=0;i<n1.len_name ;i++)
if (n1.name[i]>=65 && n1.name [i]<=90)
n1.name [i]+=32;
for (i=0;i<n2.len_name ;i++)
if (n2.name[i]>=65 && n2.name [i]<=90)
n2.name [i]+=32;
die1=0;
for (i=0;i<n1.len_name ;i++)
{
if ((n1.name [i]>=65 && n1.name[i] <=90)||(n1.name[i]>=97 && n1.name[i]<=122))
{
die1=1;
break;
}
}
die2=0;
for (i=0;i<n2.len_name ;i++)
{
if ((n2.name [i]>=65 && n2.name[i] <=90)||(n2.name[i]>=97 && n2.name[i]<=122))
{
die2=1;
break;
}
}
n1.n_value =0;
for (i=0;i<n1.len_name ;i++)
{
if (n1.name[i]>=97 && n1.name[i]<=122)
n1.n_value +=n1.name[i]-96;
}
n2.n_value =0;
for (i=0;i<n2.len_name ;i++)
{
if (n2.name[i]>=97 && n2.name[i]<=122)
n2.n_value +=n2.name[i]-96;
}

for (i=0;i<=5;i++)
{
if (n1.n_value >=10 && n1.n_value <=99)
{
n1.n_value =n1.n_value/10+n1.n_value % 10;
}
if (n1.n_value >=100 && n1.n_value <=600)
{
n1.n_value =n1.n_value /100+((n1.n_value /10) % 10)+n1.n_value % 100;
}

if (n2.n_value >=10 && n2.n_value <=99)
{
n2.n_value =n2.n_value/10+n2.n_value % 10;
}
if (n2.n_value >=100 && n2.n_value <=600)
{
n2.n_value =n2.n_value /100+((n2.n_value /10) % 10)+n2.n_value % 100;
}

}
if (n1.n_value >n2.n_value )
ans=n2.n_value /(double)n1.n_value;
else
ans=n1.n_value /(double)n2.n_value;

ans*=100;

if (die1==0 || die2==0)
printf("0.00 ");
else
printf("%0.2f ",ans);
printf("%%\n");
}
}




[/cpp]

if the ratio of the result is more than 1, say x, i take 1/x...
if user only input rubbish charactar in one of the name...
the result will be 0.00 %....

am i right?

any other rules that i havent consider?

i got a WA..
pls help me to take a look.

[Adil]

hello.

the problem-statement states that only the alphabetic characters have values. so if you get any other character, then ASCII code for that character should NOT be added.

if you haven't got AC yet, hope this helps.

[Calvin]

What a complicated program and so long for such a problem!! You should start again and think in a simple way..

For ratio, the problem is for 2 numbers A and B to calculate (A/B)*100 si A is smaller than B and (B/A)*100 if B smaller than A...[/java]

[Syed Mubashshir Husain]

The code is so complicated. Think it so easily.
The ratio

if(a > b) b / a;
else a / b;