HDU 5015 233 matrix Rapid power

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 5015

Give you a matrix of N * m (n <= 10, m <= 10 ^ 9). The elements in the first line are 233,2333, 23333 ...., the element in the first column is a given known element. Evaluate a [n] [m], A [I] [J] = A [I-1] [J] + A [I] [J-1];

Solution: Because N is small, we can push a column by column, that is, the Matrix Recursion is as follows:

/** This question is an N * M matrix (M is relatively large) A [0] [1] = 233 A [0] [2] = 2333... A [1] [0] a [1] [1] A [2] [0] a [2] [1] A [3] [0] a [3] [1] a [4] [0] a [4] [1] .. {10 0 0 0 1 }. {1 1 0 0 0} {2333, a [1] [1], a [2] [1], a [3] [1],..., 3} = {233, a [1] [0], a [2] [0], a [3] [0],..., 3} * {1 1 1 0 0}; {1 1 1 1 0} {0 0 0 1 }**/

The Code is as follows:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long longint n,m;struct mat{    LL ma[12][12];};mat init(mat x){    memset(x.ma,0,sizeof(x.ma));    x.ma[0][0]=10;    x.ma[0][n+1]=1;    for(int i=1;i<=n;i++)        for(int j=0;j<=i;j++)        x.ma[i][j]=1;    x.ma[n+1][n+1]=1;    return x;}mat cal(mat a,mat b){    mat c;    memset(c.ma,0,sizeof(c.ma));    for(int i=0;i<=(n+1);i++)    for(int k=0;k<=(n+1);k++)    for(int j=0;j<=(n+1);j++)    c.ma[i][j]=(c.ma[i][j]+a.ma[i][k]*b.ma[k][j])%10000007;    return c;}mat pow(mat a,int k){    mat e;    memset(e.ma,0,sizeof(e.ma));    for(int i=0;i<=n+1;i++)        e.ma[i][i]=1;    while(k>0)    {        if(k&1) e=cal(e,a);        a=cal(a,a);        k>>=1;    }    return e;}int main(){    int a[13];    while(~scanf("%d%d",&n,&m))    {        a[0]=233;        for(int i=1;i<=n;i++)            scanf("%I64d",&a[i]);        a[n+1]=3;        mat x;        x=init(x);        x=pow(x,m);        LL ans=0;        for(int i=0;i<=n+1;i++)            ans=(ans+x.ma[n][i]*a[i])%10000007;        printf("%I64d\n",ans);    }    return 0;}

HDU 5015 233 matrix Rapid power