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:
    PHP教程:PHPUnit学习笔记(三)测试方法进阶
    PHP教程:PHPUnit学习笔记(二)PHPUnit基本用法
    PHP教程:PHPUnit学习笔记(一)PHPUnit介绍及安装
    phpunit框架安装
    防注入(url)
    ssh 密钥详解
    JS判断登陆端是PC还是手机
    字节换算器
    gif 图片制作和拆解
    让你的网站秒开 为IIS启用“内容过期”
  • 原文地址:https://www.cnblogs.com/yuyixingkong/p/4343064.html
Copyright © 2011-2022 走看看