HDU2604 Queuing(矩阵快速幂模板)

Queuing

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4348    Accepted Submission(s): 1925

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 84 74 8

 

Sample Output

621

 

Author

WhereIsHeroFrom

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 思路：假设L长度的最后一位是m，那么只需要考虑前n-1位满足条件的个数，也就是f(n-1),如果最后一位不是m，再往前考虑一位，

那么最后三位可能是mmf,fmf,fff,mff,其中mmf,mff符合条件，mmf只需要前n-3位满足，mff只有mmff满足，即需要前n-4位满足，也

就是f(n) = f(n-1)+f(n-3)+f(n-4)。按照这样来建立构造矩阵。

#include <stdio.h>#include <string.h>#include <algorithm>#define ll long long int#define maxn 10using namespace std;int L, MOD;struct node{    int line, cal;    ll mxt[maxn][maxn];    node(){        memset(mxt, 0, sizeof mxt);    }    void init(ll v){        for(int i = 0;i <= maxn;i++)            mxt[i][i] = v;    }};node operator * (node a, node b){    node c;    c.line = a.line, c.cal = b.cal;    for(ll i = 0;i < a.line;i++){        for(ll j = 0;j < b.cal;j++){            c.mxt[i][j] = 0;            for(ll k = 0;k < a.cal;k++){                c.mxt[i][j] += (a.mxt[i][k] * b.mxt[k][j])%MOD;                c.mxt[i][j] %= MOD;            }        }    }    return c;}node operator ^ (node a, ll n){    node c;    c.init(1);    c.line = a.line, c.cal = a.cal;    while(n){        if(n&1) c = a * c;        a = a * a;        n >>= 1;    }    return c;}int main(){    node a, b, c;    a.line = 4, a.cal = 1;    a.mxt[0][0] = 9, a.mxt[1][0] = 6, a.mxt[2][0] = 4, a.mxt[3][0] = 2;    b.line = 4, b.cal = 4;    b.mxt[0][0] = b.mxt[0][2] = b.mxt[0][3] = b.mxt[1][0] = b.mxt[2][1] = b.mxt[3][2] = 1;    c.line = 4, c.cal = 4;    while(~scanf("%d %d", &L, &MOD)){        if(L == 0) printf("0\n");        else if(L <= 4) printf("%lld\n", a.mxt[4-L][0]%MOD);        else{            c = b ^ (L - 4);            c = c * a;            printf("%lld\n", c.mxt[0][0]%MOD);        }    }}