This guy is new on codechef . You guys could have told him this without the downvotes. I dont think this is a good way to welcome somone to the community.
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Lol . That was not discussion bro . He was highly ecstatic when his solution of CHEFTET was accepted within 10 mins of solving the 2nd question , and he told me the approach , and said how dumb the question was and unnecessarily the language was made hard . I had seen the question the previous day and had another approach in mind , and that is why when he told me the approach , I figured out several test cases where that solution is bound to fail 
And now he is mad at me xD
2 Likes
Well, yeah, the cases are really weak. A two line solution of mine passed.
After 3 days of this post Codechef should have done something, but still I cannot see any changes in the problem statement or rejudgement of solutions.
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Hahahhaa 
You donāt have to wait too long . 5 more days to go xD
For partial points , the test cases may be weak . OK , many will be fine with it , but for 100 ??? Are you kidding me? -_-
My solution assumes we have to use all elements of b and passes.
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So, can we use elements of b multiple times?
thanks rajan for informationā¦
but why not codechef clear this according to question B will fall on A in this scenerio one element of B can fall on unique element of
A so that element of B canāt be reusedā¦
such type of information less question kills time this is really irritatingā¦
So , is your roommatesās solution still accepted ?
no ā¦ not yet
no u need to use each element only onceā¦some subtasks failed because of that
no all B elements must fallā¦ that is why i got wrong too
by telling the complexity, you kinda revealed the entire algorithm
if(sum % n == 0 && (sum/n == a[0] || sum/n == a[0] + b[0] || sum/n == a[0] + b[1] || sum/n == a[0] + b[0] +b[1])) printf("%lld\n",sum/n);
else printf("-1\n");
^these were those 2 lines, which got 100.
https://www.codechef.com/viewsolution/10754867
This was the 2 lines of code of my roommate giving 100 