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 ai,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?
InputThere 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).OutputFor each case, output a n,m mod 10000007.Sample Input
1 1 1 2 2 0 0 3 7 23 47 16
Sample Output
234 2799 72937
Hint
矩阵快速幂
比较裸 ，首先找到原态和现态的关系，233，2333，23333，23333， 这些数都遵循 a0=a0‘*10+3 ，
a1=a0‘*10+3+a1‘ ; a2=a0‘*10+3+a1‘+a2‘ ; a3=a0‘*10+3+a1‘+a2‘+a3‘ ;
即 an=a0‘*10+3+
代码如下：
#include<iostream> #include<stdio.h> #include<cstring> const int MOD=10000007; using namespace std; typedef long long LL; const int M=15; struct Matrix{ LL matrix[M][M]; }; int n;//矩阵的阶数 void init(Matrix &res){ for(int i=0;i<=n;i++) { for(int j=0;j<=n;j++) res.matrix[i][j]=0; res.matrix[i][i]=1; } } Matrix multiplicative(Matrix a,Matrix b){ Matrix res; memset(res.matrix,0,sizeof(res.matrix)); for(int i = 0 ; i < n ; i++) for(int j = 0 ; j < n ; j++) for(int k = 0 ; k < n ; k++) res.matrix[i][j] =(res.matrix[i][j]%MOD+a.matrix[i][k]%MOD*b.matrix[k][j]%MOD)%MOD; return res; } Matrix pow(Matrix mx,int m){ Matrix res,base=mx; init(res); //初始为单位矩阵，即除主对角线都是1外，其他都是0 while(m) { if(m&1) res=multiplicative(res,base); base=multiplicative(base,base); m>>=1; } return res; } void lk_init(Matrix &b) { for(int i=0;i<n-1;i++) { b.matrix[i][0]=10; b.matrix[i][n-1]=1; } b.matrix[n-1][n-1]=1; for(int i=1;i<n-1;i++) for(int j=1;j<=i;j++) b.matrix[i][j]=1; } int main() { int m,a[M]; while(~scanf("%d%lld",&n,&m)) { for(int i=1;i<=n;i++) scanf("%lld",&a[i]); a[0]=23; a[n+1]=3; n=n+2; Matrix base={0}; lk_init(base); base=pow(base,m); LL ans=0; for(int i=0;i<n;i++) ans=(ans%MOD+base.matrix[n-2][i]%MOD*a[i]%MOD+MOD)%MOD; printf("%lld\n",ans); } return 0; }