Discuss SRM 490 there.
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ans = abs( N - gcd( N, M ) ) / 2.00 ???
Waiting time of starship is a divisor of D, because it is always linear combination of N and M.
Assume waiting time of i-th starship is ai.
ai + 1 = ( - i * M)modN, a0 = aN = 0, , so the sequence is linear and has a period. The answer is an average value of one period of the sequence.
D = x * N + y * M for some x, y (extended euclidian algorithm ), so ay = N - D, hence every k * D , 0 < = k * D < N is found in the sequence.
The answer is average of k * D where 0 < = k * D < N which is (N - D) / 2.0
Or Is there any other approach to this question?
check this link & its 2nd page!