原題鏈接

描述

Queues and Priority Queues are data structures which are known to most computer scientists. The Queue occurs often in our daily life. There are many people lined up at the lunch time. Now we define that ‘f’ is short for female and ‘m’ is short for male. If the queue’s length is L, then there are 2L numbers of queues. For example, if L = 2, then they are ff, mm, fm, mf . If there exists a subqueue as fmf or fff, we call it O-queue else it is a E-queue.
Your task is to calculate the number of E-queues mod M with length L by writing a program.

輸入

Input a length L (0 <= L <= 10e6) and M.

輸出

Output K mod M(1 <= M <= 30) where K is the number of E-queues with length L.

樣例輸入

3 8
4 7
4 8

樣例輸出

6
2
1

思路

首先是數(shù)學(xué)上找出遞推關(guān)系式$ a_{n} = a_{n-1} + a_{n-3} + a_{n-4}, n > 4 $
接下來當(dāng)然可以去解通項，不過略麻煩而且不好算，故直接構(gòu)建矩陣，用矩陣快速冪。
\[ \left[ \begin{matrix} a_{n} & a_{n-1} & a_{n-2} & a_{n-3} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right] = \left[ \begin{matrix} a_{n-1} & a_{n-2} & a_{n-3} & a_{n-4} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right] \left[ \begin{matrix} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{matrix} \right] \]

代碼

#include <cstdio> #include<cstring> #define ll long long #define maxn 4 using namespace std;int k, mod;struct Mat {ll f[maxn][maxn];void cls(){memset(f, 0, sizeof(f));}//全部置為0 Mat() {cls();}friend Mat operator * (Mat a, Mat b){Mat res;for(int i = 0; i < maxn; i++) for(int j = 0; j < maxn; j++)for(int k = 0; k < maxn; k++)(res.f[i][j] += a.f[i][k] * b.f[k][j]) %= mod;return res;} };Mat quick_pow(Mat a) { Mat ans;for(int i = 0; i < maxn; i++) ans.f[i][i] = 1;int b = k;while(b != 0) {if(b & 1) ans = ans * a;b >>= 1;a = a * a;}return ans; }int main() {Mat A, B;B.f[0][0] = 9; B.f[0][1] = 6; B.f[0][2] = 4; B.f[0][3] = 2;A.f[0][0] = A.f[0][1] = A.f[1][2] = A.f[2][0] = A.f[2][3] = A.f[3][0] = 1;while(~scanf("%d %d", &k, &mod)){Mat C;if(k <= 4) {printf("%d\n", B.f[0][4-k] % mod); continue;}k -= 4;C = quick_pow(A);C = B * C;printf("%d\n", C.f[0][0]);}return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/HackHarry/p/8391899.html

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