zoukankan      html  css  js  c++  java
  • hdu 5015 233 Matrix (矩阵高速幂)

    233 Matrix

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 749    Accepted Submission(s): 453


    Problem Description
    In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333... (it means a0,1 = 233,a0,2 = 2333,a0,3 = 23333...) Besides, in 233 matrix, we got ai,j = ai-1,j +ai,j-1( i,j ≠ 0). Now you have known a1,0,a2,0,...,an,0, could you tell me an,m in the 233 matrix?

     

    Input
    There are multiple test cases. Please process till EOF.

    For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 109). The second line contains n integers, a1,0,a2,0,...,an,0(0 ≤ ai,0 < 231).
     

    Output
    For each case, output an,m mod 10000007.
     

    Sample Input
    1 1 1 2 2 0 0 3 7 23 47 16
     

    Sample Output
    234 2799 72937
    Hint

    思路:

    第一列元素为:

    0

    a1

    a2

    a3

    a4

    转化为:

    23

    a1

    a2

    a3

    a4

    3

    则第二列为:

    23*10+3

    23*10+3+a1

    23*10+3+a1+a2

    23*10+3+a1+a2+a3

    23*10+3+a1+a2+a3+a4

    3

    依据前后两列的递推关系,有等式可得矩阵A的元素为   

    #include"iostream"
    #include"stdio.h"
    #include"string.h"
    #include"algorithm"
    #include"queue"
    #include"vector"
    using namespace std;
    #define N 15
    #define LL __int64
    const int mod=10000007;
    int n;
    int b[N];
    struct Mat
    {
        LL mat[N][N];
    }a,ans;
    Mat operator*(Mat a,Mat b)
    {
        int i,j,k;
        Mat c;
        memset(c.mat,0,sizeof(c.mat));
        for(i=0; i<=n+1; i++)
        {
            for(j=0; j<=n+1; j++)
            {
                c.mat[i][j]=0;
                for(k=0; k<=n+1; k++)
                {
                    if(a.mat[i][k]&&b.mat[k][j])
                    {
                        c.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
                        c.mat[i][j]%=mod;
                    }
                }
            }
        }
        return c;
    }
    void mult(int k)
    {
        int i;
        memset(ans.mat,0,sizeof(ans.mat));
        for(i=0;i<=n+1;i++)
            ans.mat[i][i]=1;
        while(k)
        {
            if(k&1)
                ans=ans*a;
            k>>=1;
            a=a*a;
        }
    }
    void inti()
    {
        int i,j;
        b[0]=23;
        b[n+1]=3;
        for(i=1; i<=n; i++)
            scanf("%d",&b[i]);
        memset(a.mat,0,sizeof(a.mat));
        for(i=0; i<=n; i++)
        {
            a.mat[i][0]=10;
            a.mat[i][n+1]=1;
        }
        a.mat[n+1][n+1]=1;
        for(i=1; i<n+1; i++)
        {
            for(j=1; j<=i; j++)
            {
                a.mat[i][j]=1;
            }
        }
    }
    int main()
    {
        int i,m;
        while(scanf("%d%d",&n,&m)!=-1)
        {
            inti();
            mult(m);
            LL s=0;
            for(i=0;i<=n+1;i++)
                s=(s+(ans.mat[n][i]*b[i])%mod)%mod;
            printf("%I64d
    ",s);
        }
        return 0;
    }
    




  • 相关阅读:
    2017年年终总结
    7只老鼠测试100个瓶子
    jgs--多线程和synchronized
    springboot&&vue简单的景点信息管理系统
    springboot&&vue前后端分离入门案例
    npm安装教程
    springboot整合mybatisplus
    spring整合Mybatis-plus
    Springboot简单练手的记账本
    SpringBoot整合thymeleaf简单的CRUD
  • 原文地址:https://www.cnblogs.com/cxchanpin/p/7147890.html
Copyright © 2011-2022 走看看