SDAU dp专题 1009

1：问题描述
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

2：思路

这个题找到递推公式后，要利用矩阵快速幂。
3：感想

挺难的一个题，查的题解~~
要好好研究研究。

4：ac代码

#include <cstdlib>#include <cstring>#include <cstdio>#include <iostream>using namespace std;int N,m;struct matrix{       int a[4][4];}origin,res;matrix multiply(matrix x,matrix y){       matrix temp;       memset(temp.a,0,sizeof(temp.a));       for(int i=0;i<4;i++)       {               for(int j=0;j<4;j++)               {                       for(int k=0;k<4;k++)                       {                               temp.a[i][j]+=(x.a[i][k]*y.a[k][j])%m;                               temp.a[i][j]%=m;                       }               }       }       return temp;}void calc(int n){     while(n)     {             if(n&1)                    res=multiply(res,origin);             origin=multiply(origin,origin);             n>>=1;     }}int main(){    int ff[5]={9,6,4,2};    int i,j;    matrix aaa;    //freopen("r.txt","r",stdin);    while(~scanf("%d%d",&N,&m))    {        if(N==0)        {            cout<<0<<endl;            continue;        }        if(N<=4)        {            cout<<ff[4-N]%m<<endl;            continue;        }        int doudou[5]={0};        memset(origin.a,0,sizeof(origin.a));        memset(aaa.a,0,sizeof(aaa.a));        aaa.a[0][0]=9;        aaa.a[1][0]=6;        aaa.a[2][0]=4;        aaa.a[3][0]=2;        origin.a[0][0]=origin.a[0][2]=origin.a[0][3]=origin.a[1][0]=origin.a[2][1]=origin.a[3][2]=1;        res.a[0][0]=res.a[1][1]=res.a[2][2]=res.a[3][3]=1;        calc(N-4);        res=multiply(res,aaa);        cout<<res.a[0][0]%m<<endl;    }    return 0;}