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[],sum[]; struct Matrix
{
lll m[][];
Matrix()
{
memset(m,,sizeof(m));
for(int i=;i<=n+;i++)
m[i][i] = 1LL;
}
}; Matrix Mul(Matrix a,Matrix b)
{
Matrix res;
int i,j,k;
for(i=;i<=n+;i++)
{
for(j=;j<=n+;j++)
{
res.m[i][j] = ;
for(k=;k<=n+;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&)
res = Mul(res,a);
a = Mul(a,a);
b >>= ;
}
return res;
} int main()
{
int i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
sum[] = ;
for(i=;i<=n;i++)
{
scanf("%I64d",&a[i]);
sum[i] = (sum[i-] + a[i]);
}
lll suma = sum[n];
if(m == )
{
printf("%I64d\n",(233LL+suma)%Mod);
continue;
}
Matrix base;
memset(base.m,,sizeof(base.m));
for(i=;i<=n+;i++)
base.m[i][] = 10LL;
for(i=;i<=n+;i++)
{
for(j=;j<=n+;j++)
{
if(i >= j)
base.m[i][j] = 1LL;
}
}
for(i=;i<=n+;i++)
base.m[i][n+] = 1LL;
Matrix Right;
memset(Right.m,,sizeof(Right.m));
Right.m[][] = 233LL;
for(i=;i<=n+;i++)
Right.m[i][] = (233LL+sum[i-])%Mod;
Right.m[n+][] = 3LL;
Matrix ans = fastm(base,m-);
ans = Mul(ans,Right);
printf("%I64d\n",ans.m[n+][]%Mod);
}
return ;
}