t*s
发帖数: 28 | 1 // Find the winner in the circle of N (nPlayers)
// players skipping M (nSkip) and removing Mth
// player. The winner is the last one left.
//
public int FindWinner(int a[]
int nPlayers,
int nSkip
)
其实就是那个什么joesph 问题。有什么好方法可以解？如果使用i=(i+1)%n 的方法，
complexity 不好。 | I*********g
发帖数: 93 | | t*s
发帖数: 28 | 3 google 到一个。不知道什么意思。
r := 0;
for i from 1 to n do
r := (r + k) mod i;
return r;
谁能给解释一下？ | i***h
发帖数: 12655 | 4 hmm, thought of this,
but in later round, the skip may make of more than 1 round
of the original circle.
anyone can tell me if this is really the correct solution? I doubt it
【在 t*s 的大作中提到】
: google 到一个。不知道什么意思。
: r := 0;
: for i from 1 to n do
: r := (r + k) mod i;
: return r;
: 谁能给解释一下？
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