233 Matrix（hdu5015 矩阵）

最新推荐文章于 2024-07-31 10:00:00 发布

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最新推荐文章于 2024-07-31 10:00:00 发布

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原文链接：http://www.cnblogs.com/yuyixingkong/p/4343064.html

本文探讨了在解决矩阵运算与模数计算问题时的策略与技巧，通过实例解析了如何利用矩阵快速幂求解特定问题，并强调了在处理大量数据时的效率优化。文中还详细介绍了矩阵乘法的实现方式以及如何通过快速幂技术加速计算过程，旨在为程序员提供高效解决实际问题的方法。

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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 a _0,1 = 233,a _0,2 = 2333,a _0,3 = 23333...) Besides, in 233 matrix, we got a _i,j = a _i-1,j +a _i,j-1( i,j ≠ 0). Now you have known a _1,0,a _2,0,...,a _n,0, could you tell me a _n,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 ≤ 10 ⁹). The second line contains n integers, a _1,0,a _2,0,...,a _n,0(0 ≤ a _i,0 < 2 ³¹).

Output

For each case, output a _n,m mod 10000007.

Sample Input

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

...

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\n",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！！！

转载于:https://www.cnblogs.com/yuyixingkong/p/4343064.html