HDU 5015 233 Matrix --矩阵快速幂

题意：给出矩阵的第0行（233,2333,23333,...）和第0列a1,a2,...an（n<=10，m<=10^9），给出式子： A[i][j] = A[i-1][j] + A[i][j-1]，要求A[n][m]。

解法：看到n<=10和m<=10^9 应该对矩阵有些想法，现在我们假设要求A[a][b],则A[a][b] = A[a][b-1] + A[a-1][b] = A[a][b-1] + A[a-1][b-1] + A[a-2][b] = ...

这样相当于右图：，红色部分为绿色部分之和，而顶上的绿色部分很好求，左边的绿色部分(最多10个)其实就是：A[1][m-1],A[2][m-1]..A[n][m-1],即对每个1<=i<=n, A[i][m]都可由A[1][m-1],A[2][m-1]..A[n][m-1]，于是建立12*12的矩阵：

，将中间矩阵求m-1次幂，与右边[A[0][1],A[1][1]..A[n][1],3]^T相乘，结果就可以得出了。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#define Mod 10000007

#define SMod Mod

#define lll __int64

using namespace std;



int n,m;

lll a[100006],sum[100005];



struct Matrix

{

    lll m[13][13];

    Matrix()

    {

        memset(m,0,sizeof(m));

        for(int i=1;i<=n+2;i++)

            m[i][i] = 1LL;

    }

};



Matrix Mul(Matrix a,Matrix b)

{

    Matrix res;

    int i,j,k;

    for(i=1;i<=n+2;i++)

    {

        for(j=1;j<=n+2;j++)

        {

            res.m[i][j] = 0;

            for(k=1;k<=n+2;k++)

                res.m[i][j] = (res.m[i][j]+(a.m[i][k]*b.m[k][j])%SMod + SMod)%SMod;

        }

    }

    return res;

}



Matrix fastm(Matrix a,int b)

{

    Matrix res;

    while(b)

    {

        if(b&1)

            res = Mul(res,a);

        a = Mul(a,a);

        b >>= 1;

    }

    return res;

}



int main()

{

    int i,j;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        sum[0] = 0;

        for(i=1;i<=n;i++)

        {

            scanf("%I64d",&a[i]);

            sum[i] = (sum[i-1] + a[i]);

        }

        lll suma = sum[n];

        if(m == 1)

        {

            printf("%I64d\n",(233LL+suma)%Mod);

            continue;

        }

        Matrix base;

        memset(base.m,0,sizeof(base.m));

        for(i=1;i<=n+1;i++)

            base.m[i][1] = 10LL;

        for(i=2;i<=n+1;i++)

        {

            for(j=2;j<=n+1;j++)

            {

                if(i >= j)

                    base.m[i][j] = 1LL;

            }

        }

        for(i=1;i<=n+2;i++)

            base.m[i][n+2] = 1LL;

        Matrix Right;

        memset(Right.m,0,sizeof(Right.m));

        Right.m[1][1] = 233LL;

        for(i=2;i<=n+1;i++)

            Right.m[i][1] = (233LL+sum[i-1])%Mod;

        Right.m[n+2][1] = 3LL;

        Matrix ans = fastm(base,m-1);

        ans = Mul(ans,Right);

        printf("%I64d\n",ans.m[n+1][1]%Mod);

    }

    return 0;

}

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