zoukankan      html  css  js  c++  java
  • 233 Matrix(hdu5015 矩阵)

    233 Matrix

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 1190    Accepted Submission(s): 700


    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

     

    我们这样看:已知a11 ,a21 ,a31 ,a41  。。。求后面的

    a12 = a11 +233;

    a22 = a11 + a21 +233;

    a32 = a11 + a21 +a31 +233;

    a42 = a11 + a21 +a31 +a41 +233;

    .........

    同理:后面的列也一样:

    a13 = a12 +233;

    a23 = a12 + a22 +233;

    a33 = a12 + a22 +a32 +233;

    a43 = a12 + a22 +a32 +a42 +233;

    ...........

    ss所以有矩阵:

    233 a11 
    a21  a31  a41  ... 3

    *

    10 1 1 1 1 ... 0
    0 1 1 1 1 ... 0
    0 0 1 1 1 ... 0
    0 0 0 1 1 ... 0
    0 0 0 0 1 ... 0
    ... ... ... ... ... ... ...
    1 0 0 0 0 ... 1

    =

    ......................................................................................................................................................

     

     

    z转载请注明出处:寻找&星空の孩子

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5015 

    #include<cstdlib>
    #include<cstring>
    #include<cstdio>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define LL __int64
    #define mod 10000007
    
    LL N,M;
    
    struct matrix
    {
        LL m[15][15];
    };
    LL a[15];
    
    matrix multiply(matrix x,matrix y)
    {
        matrix temp;
        memset(temp.m,0,sizeof(temp.m));
        for(int i=0; i<N+2; i++)
        {
            for(int j=0; j<N+2; j++)
            {
                if(x.m[i][j]==0) continue;
                for(int k=0; k<N+2; k++)
                {
                    if(y.m[j][k]==0) continue;
                    temp.m[i][k]+=x.m[i][j]*y.m[j][k]%mod;
                    temp.m[i][k]%=mod;
                }
            }
        }
        return temp;
    }
    
    matrix quickmod(matrix a,LL n)
    {
        matrix res;
        memset(res.m,0,sizeof(res.m));
        for(int i=0;i<N+2;i++) res.m[i][i]=1;
        while(n)
        {
            if(n&1)
                res=multiply(res,a);
            n>>=1;
            a=multiply(a,a);
        }
        return res;
    }
    int main()
    {
        int n,k;
        while(scanf("%d%d",&N,&M)!=EOF)
        {
            a[0]=233;
            a[N+1]=3;
            for(int i=1;i<=N;i++)
            {
                scanf("%d",&a[i]);
            }
    
            matrix ans;
            memset(ans.m,0,sizeof(ans.m));
            ans.m[0][0]=10;
            ans.m[N+1][0]=1;
            ans.m[N+1][N+1]=1;
            for(int j=1;j<=N;j++)
            {
                for(int i=0;i<=j;i++)
                {
                    ans.m[i][j]=1;
                }
            }
    
            ans=quickmod(ans,M);//M次幂定位到纵坐标。
    
            LL ant=0;
            for(int i=0;i<N+2;i++)//横坐标是N,即,乘以矩阵的N列。
            {
                ant=(ant+a[i]*ans.m[i][N])%mod;
            }
            printf("%I64d
    ",ant);
        }
        return 0;
    }

     

    本来要做新题的,可是遇到不会的了。。。hdu4767 Bell 现在卡在  中国剩余定理,还要好好梳理梳理!

    加油!少年!!!                                                 

  • 相关阅读:
    牛客(47)求1+2+3+...+n
    牛客(48)不用加减乘除做加法
    【推荐】可编程的热键 AutoHotkey
    【Web】js 简单动画,犯了低级错误
    【分享】JDK8u241 win x64度盘下载
    【Web】开始学Web开发!
    【数组】深析 “数组名称”
    【基础向】浅析 "多(二)维数组" 的三种引用方法
    【一个小错误】通过数组指针引用数组成员
    【网络通信教程】windows 下的 socket API 编程(TCP协议)
  • 原文地址:https://www.cnblogs.com/yuyixingkong/p/4343064.html
Copyright © 2011-2022 走看看