233 Matrix HDU - 5015

​​http://acm.hdu.edu.cn/showproblem.php?pid=5015​​

行数很小 可以想到快速幂

a[i][j]=a[i][j-1]+a[i-1][j]=a[i][j-1]+a[i-1][j-1]+a[i-2][j-1]=a[i][j-1]+a[i-1][j-1]+...+a[1][j-1]+a[0][j]

这样就找出第j列和第(j-1)列的递推关系 构造矩阵(n=5)

1 0 0 0 0

1 1 0 0 0

1 1 1 0 0

1 1 1 10 0

0 0 0 1 1

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=10000007;

ll mat[20][20],res[20][20];
ll a[20];
int n,m;

void getmul(ll u[][20],ll v[][20])
{
    ll t[20][20];
    int i,j,k;
    for(i=1;i<=n+2;i++)
    {
        for(j=1;j<=n+2;j++)
        {
            t[i][j]=0;
            for(k=1;k<=n+2;k++)
            {
                t[i][j]=(t[i][j]+(u[i][k]*v[k][j])%mod)%mod;
            }
        }
    }
    memcpy(u,t,sizeof(t));
}

void quickpow()
{
    int i,j;
    memset(mat,0,sizeof(mat));
    for(i=1;i<=n+1;i++)
    {
        for(j=1;j<=i;j++)
        {
            mat[i][j]=1ll;
        }
    }
    mat[n+1][n+1]=10ll,mat[n+2][n+1]=1ll,mat[n+2][n+2]=1ll;
    memset(res,0,sizeof(res));
    for(i=1;i<=n+2;i++) res[i][i]=1ll;
    while(m>0)
    {
        if(m%2) getmul(res,mat);
        getmul(mat,mat);
        m/=2;
    }
}

int main()
{
    ll ans;
    int i,j;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=1;i<=n;i++) scanf("%lld",&a[i]);
        quickpow();
        ans=0ll;
        for(i=1;i<=n;i++) ans=(ans+(a[i]*res[n-i+1][1])%mod)%mod;
        ans=(ans+(233ll*res[n+1][1])%mod)%mod;
        ans=(ans+(3ll*res[n+2][1])%mod)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}