Massive Algorithms: HDU 2604 - Queuing (矩阵快速幂)

Problem Description

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 2^L 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.

Input

Input a length L (0 <= L <= 10 ⁶) and M.

Output

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

Sample Input


3 8
4 7
4 8

Sample Output


6
2
1

http://www.bkjia.com/ASPjc/857172.html

题意：
n个人排队，f表示女，m表示男，包含子串‘fmf’和‘fff’的序列为O队列，否则为E队列，有多少个序列为E队列。

分析：
矩阵快速幂入门题。
下面引用巨巨解释：

用f(n)表示n个人满足条件的结果，那么如果最后一个人是m的话，那么前n-1个满足条件即可，就是f(n-1)；
如果最后一个是f那么这个还无法推出结果，那么往前再考虑一位：那么后三位可能是：mmf, fmf, mff, fff，其中fff和fmf不满足题意所以我们不考虑，但是如果是
mmf的话那么前n-3可以找满足条件的即：f(n-3)；如果是mff的话，再往前考虑一位的话只有mmff满足条件即：f(n-4)
所以f(n)=f(n-1)+f(n-3)+f(n-4)，递推会跪，可用矩阵快速幂
构造一个矩阵：

http://blog.csdn.net/chenguolinblog/article/details/10229221

思路: 递推+矩阵快速幂

分析;

1 根据题目的意思，我们可以求出F[0] = 0 , F[1] = 2 , F[2] = 4 , F[3] = 6 , F[4] = 9 , F[5] = 15

2 那么根据上面前5项我们可以求出n >= 5的时候 F[n] = F[n-1]+F[n-3]+F[n-4]

那么我们就可以构造出矩阵

| 1 0 1 1 | | F[n-1] | | F[n] |

| 1 0 0 0 | * | F[n-2] | = | F[n-1] |

| 0 1 0 0 | | F[n-3] | | F[n-2] |

| 0 0 1 0 | | F[n-4] | | F[n-3] |

int l, MOD;
struct Mat{
 ll v[SIZE][SIZE]; // value of matrix
 Mat() {
  memset(v, 0, sizeof(v));
 }
 void init(ll _v) {
  repf (i, 0, SIZE)
   v[i][i] = _v;
 }
};
Mat operator * (Mat a, Mat b) {
 Mat c;
 repf (i, 0, SIZE - 1) {
  repf (j, 0, SIZE - 1) {
   c.v[i][j] = 0;
   repf (k, 0, SIZE - 1) {
    c.v[i][j] += (a.v[i][k] * b.v[k][j]) % MOD;
    c.v[i][j] %= MOD;
   }
  }
 }
 return c;
}
Mat operator ^ (Mat a, ll k) {
 Mat c;
 c.init(1);
 while (k) {
  if (k&1) c = a * c;
  a = a * a;
  k >>= 1;
 }
 return c;
}
int main() {
 Mat a, b, c;
 // a
 a.v[0][0] = 9;
 a.v[1][0] = 6;
 a.v[2][0] = 4;
 a.v[3][0] = 2;
 // b
 b.v[0][0] = b.v[0][2] = b.v[0][3] = b.v[1][0] = b.v[2][1] = b.v[3][2] = 1;

 while (~scanf("%d%d", &l, &MOD)) {
  if (l == 0) {
   puts("0");
  } else if (l <= 4) {
   printf("%lld\n", a.v[4 - l][0] % MOD);
  } else {
   c = b ^ (l - 4);
   c = c * a;
   printf("%lld\n", c.v[0][0] % MOD);
  }
 }
 return 0;
}

Also check http://972169909-qq-com.iteye.com/blog/1857117

HDU 2604 - Queuing (矩阵快速幂)

Labels

Popular Posts