zoukankan      html  css  js  c++  java
  • HDU2604 Queuing 矩阵初识

    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. 

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

    下图是对矩阵的理解,对左边每一个f(n),需要m个f(x)就在第x排记录m。
    如图,f(n)=x*f(n-1)+y*f(n-2)+z*f(n-3)...
    本题的公式是f(n)=f(n-1)+f(n-3)+f(n-4),分别对最后一位是f或者讨论即可得出。


    #include<cstdio>
    #include<cstdlib>
    #include<algorithm>
    #include<memory>
    #include<cstring>
    #include<cmath>
    using namespace std;
    const int maxn=4;
    int Mod;
    struct mat
    {
        int m[maxn][maxn],len;
        mat(){memset(m,0,sizeof(m));len=maxn;}
        mat friend operator * (mat a,mat b){
            mat d;
            for(int i=0;i<a.len;i++)
             for(int j=0;j<a.len;j++){
              d.m[i][j]=0;
               for(int k=0;k<a.len;k++)
                d.m[i][j]+=(a.m[i][k]*b.m[k][j])%Mod;
             }
            return d;
        }
        mat friend operator^(mat a,int k) {
            mat c;
            for(int i=0;i<c.len;i++) c.m[i][i]=1;
            while(k){
               if(k&1) c=a*c;
               a=a*a;
               k>>=1;
            }
            return c;
       }
    };
    int main()
    {
        int n,k;
        mat ans,x,c;
        x.m[0][0]=x.m[0][2]=x.m[0][3]=x.m[1][0]=x.m[2][1]=x.m[3][2]=1;
        ans.m[0][0]=9;
        ans.m[1][0]=6;
        ans.m[2][0]=4;
        ans.m[3][0]=2; 
        while(~scanf("%d%d",&n,&Mod)){
            if (n<=4){
                printf("%d
    ",ans.m[4-n][0]%Mod);
            }
            else {
                c=x^(n-4);
                c=c*ans;
                printf("%d
    ",c.m[0][0]%Mod);
            }
        }
        return 0;
    }
    View Code
  • 相关阅读:
    Convolutional Neural Network-week1编程题(一步步搭建CNN模型)
    Coursera Deep Learning笔记 卷积神经网络基础
    爬取b站周杰伦新歌mv弹幕 绘制词云
    Coursera Deep Learning笔记 结构化机器学习项目 (下)
    Golang-执行go get私有库提示”410 Gone“ 解决办法
    golang常用的http请求操作
    关于asyncio知识(四)
    关于asyncio知识(二)
    Python Every Class Needs a __repr__
    关于asyncio知识(一)
  • 原文地址:https://www.cnblogs.com/hua-dong/p/7748357.html
Copyright © 2011-2022 走看看