Suppose three gunmen are A, B, and C who have a probability of 100%, 50% and
33% respectively. The shooting will start from C, then B and at last A.
Now there are several possibilities for C. If C shoots B, then A would shoot
C with an accuracy of 100% or in other case if C shoots A, then B would
@Dave: first of all By shooting in air I meant that C will not fire any
one. That was my figure of speech:))
He will simply waste his shot.
@Salil: its a duel. everyone will get chance to shoot in each round
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@snehal:
will the shooting take place in increasing order of accuracy of hitting the
target and is that at a time only one person can take a shot???
if yes then
@Salil:
my answer would be the same as above. what C will do is that it will first
let A and B kill each other first.
After C wastes his
Is there any condition that the people in the city may or may not know each
other???
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for 2nd problem:
statement inside the do-while is executed once. Now for the condition
checking inside the while loop. The condition i++5 is evaluated which turns
out to be true. So the rest of the part ++ch='F' is not evaluated and hence
letter A is printed six times. Now when the condition i++5
correct me if I am wrong.
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@prodigy: how is it coming O(nlogk) can u explain???
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Take two pointers p and q. Initially p points to head.
while(p!=given pointer)
{
p=p-link;
q=p;
};
Now you hv pointer at 3rd and 4th position. Now insertion is simpleHope
this will work
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@shobhan: Ya,I got it!!
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