jvinniec Ok, this took me the better part of an hour to figure out what they are asking in this question. Basically, the idea is you are given two numbers 'a' and 'b' such that a<=b. You then need to test every number from a to b and output either the text version of the variable (if number <=9) or whether the value is even or odd (if number > 9).
So basically when the inputs are 8,11 you need to test every integer from 8 to 11, so you need to test 8,9,10,& 11. This leads to the output specified in the "Sample Output" of "eight, nine, even, odd". Hopefully this helps someone out.
HMansolf Thank you. This really clarified things. Rewrote and got the code running in about 30 seconds after reading this.
Abhi1428 Thanx,it really helped..
cs15m059 thanks,it really Helped more
menet_ludo Thanx, it tooks me also a little bit of time before understanding what was the purpose of this exercice. It's not clear.
CHODHARY include
include
using namespace std; int main() { int a,b; cin>>a>>b; for(a;a<=b;a++) {
if(a<=9) { if(a==8) cout<<"eight\n"; else if(a==9) cout<<"nine\n"; else if(a==7) cout<<"seven\n"; else if(a==6) cout<<"six\n"; else if(a==5) cout<<"five\n"; else if(a==4) cout<<"four\n"; else if(a==3) cout<<"three\n"; else if(a==2) cout<<"two\n"; else if(a==1) cout<<"one\n"; } else {if(a%2==0) cout<<"even\n"; else cout<<"odd\n"; }} return 0; }
liordah01 its easier and better looking to use a switch statement: switch (count){ case 1: cout << "one" << endl; break; case 2: cout << "two" << endl; break; case 3: cout << "three" << endl; break; case 4: cout << "four" << endl; break; case 5: cout << "five" << endl; break; case 6: cout << "six" << endl; break; case 7: cout << "seven" << endl; break; case 8: cout << "eight" << endl; break; case 9: cout << "nine" << endl; break; default: if (count%2 == 0) cout << "even" << endl; else cout << "odd" << endl;
tpirramba exactly...but here the quesion is realted to loop so we should ans according to the question.
Tejas1996 but the questionnis looking for the numbers(in words) in order. so if you use break statement they wont outcome in order.
shuklarituraj9 This is my code
#include <iostream> #include <cstdio> using namespace std; int main() { int i,c,b; string a[11]={"even","odd","one","two","three","four","five","six","seven","eight","nine"}; cin>>c>>b; for(i=c;i<=b;i++) { if((i>9) && (i%2==0)) cout<<a[0]<<endl; else if((i>9) && (i%2!=0)) cout<<a[1]<<endl; else cout<<a[i+1]<<endl; } return 0; }
ashutoshgoswami1 nice bruh!
gajendra14695 in your code why are u useing the variable c,b;
simranapril12 thanks man i was missing somnething out. my code is the same. arrays solved it in some lines of code!
avinashak558 What if b>c?? it wont work for b>c test case
balakumaran428 that's great man, you have clarified my doubt. thanks a lot
420vijay47 thnkx
s1035111590 perfect....!
abhileshgupta74 nice code
vvyy346 you have to remove the break statements otherwise you will have only one output ( of the matching case ). if you skip the break statements you will get all the outputs after your matching statement
danielgcwik thanks this works in c++. I get problems with cin undclared if I used C
pradyumnachak98 same here
srikanthhubli c dosnt hace cin u should use scanf
bergart11 Great code, I like it.
Here is mine:
#include <iostream> #include <cstdio> using namespace std; int main() { // Complete the code. int a,b; int n=0; string intMap[9]= {"one", "two","three","four","five","six","seven","eight","nine"}; cin>>a>>b; if ((a<=9)&&(b<=9)){ for(n=a;n<=b;n++){ cout << intMap[n-1]<<endl; } } else if ((a<=9)&&(b>9)){ for(n=a;n<=9;n++){ cout<<intMap[n-1]<<endl; } for(n=10;n<=b;n++){ if (n%2==0){ cout<<"even"<<endl; } else { cout<<"odd"<<endl; } } } else { for(n=a;n<=b;n++){ if (n%2==0){ cout<<"even"<<endl; } else { cout<<"odd"<<endl; } } } return 0; }
piyushp0541 for(n=10;n<=b;n++){
if (n%2==0){ cout<<"even"<what these lines are performing can anyone explain please ....Vexify //Compares the values of a, b with 9 and if values <= 9 then //Prints all values from 'a' uptill 'b' in string format if ((a<=9)&&(b<=9)){ for(n=a;n<=b;n++){ cout << intMap[n-1]<<endl; } } //Compares the values of a, b with 9 //And if the value of a <= 9 and value of b > 9 then.... else if ((a<=9)&&(b>9)){ //Prints all the values starting from 'a' uptill 9 //in string format and then.... for(n=a;n<=9;n++){ cout<<intMap[n-1]<<endl; } //For values which are greater than 9 and <= to 'b' //Prints 'even' or 'odd' for them for(n=10;n<=b;n++){ if (n%2==0){ cout<<"even"<<endl; } else { cout<<"odd"<<endl; } } } Hope that helps.
samarthchadda_i1 why you have used intmap[n-1]
fatursetiawan80 thank you so much...
sanjay_kanakkot [deleted]sanjay_kanakkot for(n=10;n<=b;n++) checking the value of n is less than b if(n%2==0) suppose the value of n you given is 10 if statement will find the remainder by using the modulus % that is n%2 10%2 just divide the reminder will always be zero which means if its divisible by 2 it is even number..
klausskent17 you have a for loop and it start counting from 10 , after check the condition if n is inferior or egal to b , when it is true it increment the value of n ... if n%2 == 0 , means if n is divisible by 2 , he will display even I hope it helps you :-)
eshank_tiwari [deleted]abhisheksud96 what is the use of String intMap[]
vasu71051 can u please explain the string type in C++ .I searched it over net but didn't got the satisfying result. According to me String is an array of char but you used it as a predefined data-type.how?
talalawaiskhan Love your code but arrays havent been introduced here yet! so why did u use it? Cant it be done without arrays?
V0dK6 After your "else if", your "else" statement can be skipped. It will pass all tests cases without. Just saying... Your code is great by the way and pass all tests cases anyway ;-)
shuklarituraj9 Much simple, try this.
#include <iostream> #include <cstdio> using namespace std; int main() { int i,c,b; string a[11]={"even","odd","one","two","three","four","five","six","seven","eight","nine"}; cin>>c>>b; for(i=c;i<=b;i++) { if((i>9) && (i%2==0)) cout<<a[0]<<endl; else if((i>9) && (i%2!=0)) cout<<a[1]<<endl; else cout<<a[i+1]<<endl; } return 0; }
codertoaster Unnecessarily making it so long and unreadable.
string words[]={"one","two","three","four","five","six","seven","eight","nine"}; for(int n=a;n<=b;n++) { if(n>9) if(n%2==0) printf("even\n"); else printf("odd\n"); else printf("%s\n",words[n-1].c_str()); }
shuklarituraj9 I think it is better than writing multiple "if-else" and one who can code can read it without any trouble.
cdabbott This works but is not correct, it will not work if the numbers greater than 9 are fewer than the numbers that are les than or equal to 9.
for example a=3 b=5 out put would be; three four five
no odds or evens
rvinothkanna Thanks a lot
keh_mario Don't waste your time typing all that out
int main() { // Complete the code. int a, b; cin>>a>>b; string num[9] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine"}; for(int i=a; i<=b ;i++) { if(i<=9) cout<<""<
rattikarn_dudley Thanks! this helped a lot.
Adil Thank you!
akash4u thank you...
redpix_ [deleted]
alextaylor I could not make sense of the original question as written. Thank you.
bigdan4712 That's because the author made the problem over-complicated. This is supposed to be an introductroy style problem and he went way over the mark. I was able to read the problem because I have a minor in math and have studied number and set theory in discrete mathematics.
hugo_lin It's helpful, thank u
OrionH this definitely helped. cheers man
NapdaN Thanks !!!
AbhishekKrishna Thank you, helped me understand the problem. The question needs to be edited to make it more clear.
shemul [deleted]souravsam definitely it helped a lot :) tnx
geo_hackerrank thank you .it helps me a lot
furiacs Thanks man!
trailblazerr gr8
ajit4all21 //wth is wrong with this code?
include
using namespace std; int main() { int a,b; cin>>a>>b;
for(a;a<=b;a++) { if(a<=9) { if(a==8) cout<<"eight\n"; else if(a==9) cout<<"nine\n"; } else { if(a%2==0) cout<<"even\n"; else cout<<"odd\n"; } } return 0;}
xsmgxbrayley Try changing "else if(a==9)" to just an else statement because if/else chains always need to end with an else statement to default to if no other case is true.
FiyinDaniel your code would only work if one of the numbers entered was eight. you should include all numbers less than 9.
drkato That is not true. It is perfectly acceptable to have an if/if else with no last else case if your intention is to have a situation where no statement executes if your specific conditions are not met. It is analogous to writing...
else ;
maximshen [deleted]shuklarituraj9 Use this code
#include <iostream> #include <cstdio> using namespace std; int main() { int i,c,b; string a[11]={"even","odd","one","two","three","four","five","six","seven","eight","nine"}; cin>>c>>b; for(i=c;i<=b;i++) { if((i>9) && (i%2==0)) cout<<a[0]<<endl; else if((i>9) && (i%2!=0)) cout<<a[1]<<endl; else cout<<a[i+1]<<endl; } return 0; }
Khushboo_Watwani Thank you so much for explaining this question... :)
akhii int a,b,i; cin>>a; cin>>b; for(i=a;i<=b;i++) { switch(i) { case 1: cout<<"one \n";break; case 2: cout<<"two \n";break; case 3: cout<<"three \n";break; case 4: cout<<"four \n";break; case 5: cout<<"five \n";break; case 6: cout<<"six \n";break; case 7: cout<<"seven \n";break; case 8: cout<<"eight \n";break; case 9: cout<<"nine \n";break; } } for(i=10;i<=b;i++) { if((i%2)==0) cout<<"even \n"; else cout<<"odd \n"; }
decoder_ Ain't this wrong?
akhii no its correct
decoder_ what if the value I give is a=8 and b=5?
akhii You will be given two positive integers, a and b (a<=b), separated by a newline. read the question first .....AH
decoder_ calm down okay! It was just a mistake. Sorry that I asked you.
drkato Good practice would be to not assume such things, so your intuition is spot on. A simple check at the beginning of the function hurts nothing, as so...
int a, b; cin >> a; cin >> b; if((b - a) < 0) return 0; //or whatever other error handling is needed
RA1511008010245 you are right .
shahashish991 [deleted]akhii its coreect u mother fucker
andriesuaktiwa lmao the comment and reply.. hahaha you guys, made my day
b6032682 very funny stuff
deekshith565 [deleted]
shajol484 thanks
samplenhold Pretty poor solution. Unnecessary use of two for loops. You can do it in one for loop. The code would be easier to read if it was written using one loop and in the loop had an initial if/else structure that determines if the current number in the iteration is less than or equal to 9. If the count is less than 10, you have if statements that print the english word out, and if the count is greater than or equal to 10, then you do modulo to determine even or odd.
alice_ferrazzi using a double for loop is not necessary
bartlomiejpopek This is wrong. When I Try imput more then 10 eg. 23 and 25 I get to many even and odds.
shuklarituraj9 Try this
#include <iostream> #include <cstdio> using namespace std; int main() { int i,c,b; string a[11]={"even","odd","one","two","three","four","five","six","seven","eight","nine"}; cin>>c>>b; for(i=c;i<=b;i++) { if((i>9) && (i%2==0)) cout<<a[0]<<endl; else if((i>9) && (i%2!=0)) cout<<a[1]<<endl; else cout<<a[i+1]<<endl; } return 0; }
haberpeter I see this kind of solution in many cases. But in fact this is wrong. This works only in case the first input is less than the second one. If the test case would be 12 and 11 then this failed the test case because the code does not enter into the for loop, because c is greater than b.
vvyy346 bro just skill the break statement in switch case and the you will be needing only second loop
Meenachinmay but what you will do when both numbers are between 1 and 9...because two given value may be anything,,,may be both are greater than 9 than what?
Khushboo_Watwani if two given numbers are between 1 and 9...
like: 2 4
then output: two three four
and if two numbers are greater than 9...
like: 20 23
then output: even odd even
Meenachinmay try this with some standard input samples,,,,it may be more comfortable,,,,MAY BE
mukulpahwa504 for 20 23 output: even odd even odd
bindass844 yess
GrimNight Thx A lot!!!!
Meenachinmay not at all.,,,just remember me in ur wish...:)
Tushar_Mehr You the real MVP!
rushabhajain4444 Thanks..
dynamicsahithi thank you so much
dulini88 Thank you very much for help understanding the original question.
mr_seb Thank you!
givonz i have found that a significant number of problems, need a LOT better definition. the sample/input confuses the normal interpretation of the question.
givonz i have found that a significant number of problems, need a LOT better definition. the sample/input confuses the normal interpretation of the question.
givonz now i understand why this doo hickey problem was asked before having discussed arrays and; why it needs a loop!!!!!
jingren1021 thank you very much
SidPrajosh thank you... not helps someone...its helping a lot of people...
iamrizubiswas Very helpful! saved a lot of time ,Thanks :)
jitender1712 Thanks , earlier the expected output looks bizzare to me. But now , I got the problem.
xgasko Thanks
hacktik i m still a bit confused... if (no<=9) the text version of it must be got so if input is 8.. we got to get eight.. thats alrite but after that the condition goes like if(no>9) then print even or odd respective of the no.. so for input 11 we must print it as odd only right? the output must be like "eight odd" only no, how does the ouput cums like "eight nine, even,odd" ???
xgasko Yes, if the number is less or equal than 9, then print the English representation of it. If it's bigger, you should print "even" or "odd", depending on the number. Now, your input contains two numbers. You have to print every number that is between those numbers, including those two numbers. For example, input is 8 and 11 - you print:
eight
nine
even
odd
If the input is 2 and 6, you should print:
two
three
four
five
six
hacktik thanks a lot for ur kind response... got figured out in a very clear way !!
Abul_Kalam Yup!... It really works.......
ascotan I'm glad i looked at the discussion here. It took me all of 15 seconds to fix my code after reading this. The question is clear as mud.
EtsRk Thanks..it helps me a lot.
Sivakumarramar thnx a lot
ricargu2010 this assignment was not clear at all, what they should say is print from the first given number to 9 in the form of "one", "two" ... "nine", and "even" , "odd", "even", to the second given number.
smokhele48 Question is not that clear. This got me in the right track. :)
menglish2451 Great explanation of the problem, I wasn't understanding the input until now.
rheasanjan Thank you so much!!
Parashuram_Joshi Thanks, it helped me
jwc_it_studio Thanks, I didn't understand what I was messing up till I read your explination.
surajsetty_sbi thanq soo much
ojas_k30 thanks
h754548560 Thank you very much, It's help a lot.
kashif_i This really helped in explanation of the orignal question.
m_patterson Thanks. It's a really badly worded question. It's very easy when explained clearly as you've done.
muhammadburhanz1 someone please told them to correct the question
vatsal09 Thanks for this.
naiktanvi Thanks! The comment really helped.
laurap Oh my gosh thank you!
stayx thanks a lot
ssanck Thank you !
fireash Horribly worded problem
caitenkhongbitr1 Thank you a lot. It really helpful.
zwlady253 Thanks . you saved me .
Engineer_x really helpful....thanks bro.
maheshvangala191 Thank you jvinniec Your suggestion helped me to solve this problem thank you very much
kai_abhik Thanks, I was really annoyed.
ABInfinity thanks bro that was really helpful..
vishalshukla044 dont know what to say just one word thnxs a lot
snmahajan30 yeah it really helped.
naveensurya3010 thanks bro
naveensurya3010 thanks bro
paraschopra2000 Helped alot, Thanks.
chancs88 Thank you. That helps a lot.
samarthchadda_i1 thats good but not great we can have other simple alternatives
Sarques thanks so much, i was just lost in the problem. I was just don't getting the question.
vishariaraj Thank you so much. You clarified the question. :)
shreyanshgoyal90 you actually saves my time thanks....
ketansuthar899 Thank you.
zppr12 Thanks a lot sir
sparklingboy11 it helped a lot man.Thank you
rschoudhary1999 Thanks, couldn't figure it out till I saw your comment. Really appreciate it. Helps us beginners alot.
alexbryant Wow, the instructions don't say that at all.
jose_lsandra Thank you so much. I was sitting with this for 2 hours and never understood what this was about. Thanks.
madhuvarmapenme2 thanku....it really helped in understanding what they asked for!!
yvohu2002 Haha thanks!, this question was very poorly written.
imtherealtoast Yes it does! The way these questions are put is often more complex than the code needed lol.
mkathpal09 thankyou.
mohitkumar200131 thank you now i got it
olavo_o_neto Thanks, I was not understanding the question
ramyaasrih tq so much,, this cleared all my doubts
haniyeh_ghassami really thanks.
vengelke Thank you! It really helped.
tiwarisiddharth3 Thanks alotttt ......
22pradyumns thanks bud
Dineshs91 Input format and sample input are different. The question is ambiguous.
pavelkushtia Yes, no newline between the inputs. Either the description or the test case input should be corrected.
fogbank Indeed. In testcases 1-5 the numbers are separated by a space. I'm not changing my code, since it's correct for the problem definition. The mistake is in the testcases, not in my code.
rockpile int a,b; string num[10] = {"zero","one","two","three","four","five","six","seven","eight","nine"}; cin >> a >> b; for (int i = a; i <= b; ++i) { if (i > 9) { i % 2 == 0 ? cout << "even\n" : cout << "odd\n"; } else { cout << num[i] << endl; } } return 0;
AffineStructure wow I really enjoy the ternary operator.
Bharath113 [deleted]nhantnguyen233 #include <iostream> #include <cstdio> using namespace std; int main() { // Complete the code. string const number[] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine"}; int a, b; cin >> a >> b; for (int i = a; i <= b; i++) { if (i > 9) cout << (i % 2 ? "odd" : "even") << endl; else cout << number[i - 1] << endl; } return 0; }
mitutee wity method ^)
shahashish991 go fuck youtself
decoder_ why you're not worth fucking someone?
kalyannaanu666 [deleted]
ibroheem Test for -ve val (i >=1 and i <=9)
SudhanshuMohan Very simplistic. Good job.
jorgechiquinv The ternary operator is the best!
int main() { string hour[]={"even","odd","one","two","three","four","five","six","seven","eight","nine"}; int a,b; for(cin>>a>>b;a<=b;a++) cout<<hour[a>9?a%2:a+1]<<'\n'; return 0; }
shivanidalmia131 very nice
ngochungnguyen21 very good! tks for share...
ramana_cpp excellent
chancs88 beautiful
agnelpb That's awesome. Thanks.
Manbear This challenge needs editing, because the description is very vague and can only be solved if referring to the comments section for an explanation.
captainhampton The way this question is phrased is awful. This could use quite an update to make it more clear precisely what is required.
peecee Anybody else feel this was a badly framed question?
ramey_steven Garbage question. Needs to be reworded.
polmki99 Ternary operator FTW!
string nums [10] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"}; int start, end; cin >> start; cin >> end; for (int i = start; i <= end; i++) { cout << (i < 10 ? nums[i] : i % 2 == 0 ? "even" : "odd") << endl; }
___jaw___ Yours is obviously better.
cin >> start >> end; for(int i = start; i <= end; i++){ (i > 9)?((i%2) == 0)? printf("even\n"):printf("odd\n"):(i > 8) ? printf("nine\n"):(i > 7) ? printf("eight\n"):(i > 6)? printf("seven\n"):(i > 5)? printf("six\n"):(i > 4)? printf("five\n"):(i > 3)? printf("four\n"):(i > 2)? printf("three\n"):(i > 1)? printf("two\n"):printf("one\n"); }
Geoff_Fitz include
include
using namespace std;
int main() { int num1,num2; cin >> num1 >> num2; string Arr1[9] = {"one","two","three","four","five","six","seven","eight","nine"}; for(int i =num1; i <= num2;i++) { if(i <= 9) { cout << Arr1[i-1] << endl;
} else { if(i%2 ==0) { cout << "even" << endl;
} else { cout << "odd" << endl;
} } } return 0; }Tried to make it short as possible;
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