Write a love calculator that will use the FLAMES love calculation to compute the future of two partners. How the FLAMES love calculation works :
* You will be given two different names as the input, one for each of the partners.
The common letters in the name will cancel out or will be removed, that is if you are given Elvis and Lois, l, i and s will cancel out leaving Ev and O.
* The remaining letters in each name would be counted and added together then stored in a variable, the above example only three letters are left.
* Use the stored variable to go across the word FLAMES. That is you will have to go through each letter and increment a counter variable each time you pass each letter and start all over from F again until you get the number stored in your variable(the number of letters remaining). Example: 3 falls under letter A; hint: counter = 0; variable = 3 while() { counter++; if counter != variable { goto next letter on flame } else { print result } }. Based on the last letter the loop stopped at, you output their fate/future, Where F = Friends, L = Lovers, A = Admirers, M = Married, E = Enemies, S = Sister.
* Print out the two names with the corresponding value. Example "Elvis and Lois are ADMIRERS". You can also add a percentage that tells how true the result is.
RULES 1) Run your code on www.ideone.com .
2)Any programming language is allowed.
3)your code should be fast enough to run properly on a poor cpu.
4)The person with the shortest and fastest code wins.

Try to paragraph your post, its kinda difficult undastanding it.

Modified

Be more specific at your problems @OP! ...
Here are 2 optimized solutions . . ...
The close to fastest... C++

http://ideone.com/T2RerE
took 0 seconds to run..

/* (C) WhiZTiM 2013 */
#include <iostream>
#include <string>
#include <cstring>
#include <vector>

using namespace std;
typedef string::size_type szT;
typedef vector<string> VEC_string;

void strip_common_characters(string& A, string& B)
{
	//choose the shortest string to save iteration time
	string& S1 = A.size() <= B.size() ? A : B;	//by reference!
	string& S2 = A.size() > B.size() ? A : B;	//by reference!
	szT i = 0;
	while(i < S1.size())
	{
		bool found = false;
		szT lc = S2.find(S1[i]);
		if(lc != string::npos) found = true;
		else
		{
			lc = S2.find(tolower(S1[i]));
			if(lc != string::npos) found = true;
		}
		if(found)
		{
			S1.erase(i, 1);
			S2.erase(lc,1);
			continue;
		}
		++i;
	}
}

void instantiateMap(VEC_string& mp)
{
	mp.push_back("Friends";
	mp.push_back("Lovers";
	mp.push_back("Admirers";
	mp.push_back("Married";
	mp.push_back("Enemies";
	mp.push_back("Sisters";
}

void run()
{
	string s1, s2, s3, s4;
	getline(cin, s1);
	getline(cin, s2);
	s3 = s1; s4 = s2;

	VEC_string mp;
	instantiateMap(mp);

	strip_common_characters(s1, s2);
	unsigned k = s1.size() + s2.size();
	k = (k - 1) < 0 ? 0 : ((k - 1) % mp.size());

	cout << s3 << " and " << s4 << " are " << mp[k] << endl;
}

int main(int argc, char* argv[])
{
	run();
	return 0;
}

What's wrong with these smileys? everytime.


dis one na die o!!

Another... much shorter SLoC...

http://ideone.com/HKdiNS

0.01second on my PC... when I piped input to stdin
It works on my Machine... I am on a Mobile now... can't really make it run on IDEone... this is my 3rd attempt to make it work....
#annoyed.. bad phone, MTN frustrating me and my Money...

PYTHON 2.5+ < 3.0

#!/usr/bin/python
#(C) WhiZTiM 2013

def commonLen(s1, s2):
    s1 = s1.lower()
    s2 = s2.lower()
    k = 0
    while(k < len(s1)):
        if(s2.find(s1[k]) == -1):
            k += 1
            continue
        s2 = s2.replace(s1[k], '', 1)
        s1 = s1.replace(s1[k], '', 1)
    return ( len(s1) + len(s2) )

def getVerdict(s1, s2):
    mp = ["Friends", "Lovers", "Admirers", "Married", "Enemies", "Sisters"]
    ln = commonLen(s1, s2)
    ln = 0 if( (ln - 1) < 0) else ((ln - 1) % 6)
    return mp[ln]

def main():
    s1 = raw_input()
    s2 = raw_input()
    ans = s1 + " and " + s2 + " are " + getVerdict(s1, s2)
    print ans

if(__name__ == '__main__'):
    main()

#PythonSquad

@op. Thanks for the post... I remember something like this way back in Junior School... But there was something I am not sure I understood from your question . .

WhiZTiM: @op. Thanks for the post... I remember something like this way back in Junior School... But there was something I am not sure I understood from your question . .

wats that ?

elvis10ten: wats that ?

for two names KABA and BAKKE...should the occurence K in KABA cancel out the 2 Ks in BAKKE? or just one K?

A concrete cancellation example.
iNPUT: KABA and BAKKE
OUTPUT: exclusive chars are "" and "E"
or
OUTPUT: exclusive chars are "A" and "KE"

here is my pseudocode solution in Java...



Sorry, I ment Arabic

ﺎﻬﻧﺍ ... ﺓﺮﻔﺸﻟﺍ ﻒﻄﺘﻘﻣ ﺲﻴﻟ ﺍﺬﻫ !ﺍﻮﻔﻋ !ﺔﻌﺘﻤﻠﻟ ﺩﺮﺠﻣ

^^^ *sharpening cutlass*

WhiZTiM:
for two names KABA and BAKKE...should the occurence K in KABA cancel out the 2 Ks in BAKKE? or just one K?

A concrete cancellation example.
iNPUT: KABA and BAKKE
OUTPUT: exclusive chars are "" and "E"
or
OUTPUT: exclusive chars are "A" and "KE"

only one k. One letter for one letter.

Lmao, you dey find trouble abi.

WhiZTiM: here is my pseudocode solution in Java...



Sorry, I ment Arabic

ﺎﻬﻧﺍ ... ﺓﺮﻔﺸﻟﺍ ﻒﻄﺘﻘﻣ ﺲﻴﻟ ﺍﺬﻫ !ﺍﻮﻔﻋ !ﺔﻌﺘﻤﻠﻟ ﺩﺮﺠﻣ

@whiZTiM, you are the winner of this challenge. Bravo. Your solution is the best so far.