推 sb107912:用對角化解 真是漂亮 <O> 02/03 18:42
→ gskman:我覺得就每一次分裂三個出來就3^100就對了阿..只有問總各數 02/03 19:07
→ BigTora:對欸！！只有問總數的話就3^100就好，囧" 02/03 20:09
※ 引述《PrettyPingzi (Pingzi)》之銘言：
: 15.There are two kinds of particles insides a nuclear reactor. In erery
: second, an "A" particle will spilts into three "B" particles,and a "B"
: will spilts into an "A" particles and two "B" particles. If there is a
: single "A" particles in the reactor at t=0, how many particles are there
: altogether at t=100?
: 這題我對遞回的定義不知道該怎麼去訂
: 我看黃子嘉的解題書 是 A(t)=B(t-1)
: B(t)=3A(t-1)+2B(t-1)
: A(0)=1 ,B(O)=0
: B一次分裂不是分1A2B嗎??? 那上述的遞回定義怎跑出來的囧
: 還是我想的地方有問題???
: 有請板友幫忙了 感激不盡^^
因為A會變成3個B，而B會變成1A2B
t = 0假如有x個A，y個B，
t = 1就會有y個A，3x + 2y個B，
所以A(t) = B(t - 1)
B(t) = 3A(t - 1) + 2B(t - 1)
不過自己覺得這題用對角化解比較簡單 @@
┌ ┐
令A = │0 1│
│3 2│
└ ┘
100
┌ ┐ ┌ ┐
│0 1│ │1│ 就是答案了
│3 2│ │0│
└ ┘ └ ┘
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