2014西安网络赛1009||hdu5015 矩阵

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

Problem Description

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
a0,1 = 233,a0,2 = 2333,a0,3 = 23333...) Besides, in 233 matrix, we got ai,j = ai-1,j +ai,j-1( i,j ≠ 0). Now you have known a1,0,a2,0,...,an,0, could you tell
me an,m in the 233 matrix?

 

Input

There are multiple test cases. Please process till EOF.

For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 109). The second line contains n integers, a1,0,a2,0,...,an,0(0 ≤ ai,0 < 231).

 

Output

For each case, output an,m mod 10000007.

 

Sample Input

1 1
1
2 2
0 0
3 7
23 47 16

 

Sample Output

234
2799
72937

Hint



解题思路：这题关键是矩阵的构造，比赛时是队友推出的公式，代码实现如下，细节读者品味，我就不罗嗦了。。。



#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;

typedef long long LL;
const int INF=0x3f3f3f3f;
const int N=10005;
const int MOD=10000007;

int n;
LL m,a[20];
struct Matrix
{
LL m[15][15];
};
Matrix I;
Matrix mult(Matrix a,Matrix b)
{
Matrix c;
for(int i=0;i<n+2;i++)
for(int j=0;j<n+2;j++)
{
c.m[i][j]=0;
for(int k=0;k<n+2;k++)
c.m[i][j]+=(a.m[i][k]*b.m[k][j])%MOD;
c.m[i][j]%=MOD;
}
return c;
}
Matrix quick_mod(Matrix a,LL n)
{
Matrix c=I;
while(n)
{
if(n&1)
c=mult(c,a);
n>>=1;
a=mult(a,a);
}
return c;
}

int main()
{
while(~scanf("%d%I64d",&n,&m))
{
a[0]=3;
a[1]=23;
for(int i=2;i<n+2;i++)
scanf("%I64d",&a[i]);
for(int i=0;i<n+2;i++)
for(int j=0;j<n+2;j++)
if(i==j)
I.m[i][j]=1;
else
I.m[i][j]=0;
/* for(int i=0;i<n+2;i++)
{
for(int j=0;j<n+2;j++)
printf("%I64d ",I.m[i][j]);
printf("\n");
}*/
Matrix A;
for(int i=0;i<n+2;i++)
for(int j=0;j<n+2;j++)
if(i>=j)
A.m[i][j]=1;
else
A.m[i][j]=0;
for(int i=1;i<n+2;i++)
A.m[i][1]=10;
LL ans=0;
Matrix B=quick_mod(A,m);
/*for(int i=0;i<n+2;i++)
{
for(int j=0;j<n+2;j++)
printf("%I64d ",B.m[i][j]);
printf("\n");
}*/
for(int i=0;i<n+2;i++)
ans=(ans+B.m[n+1][i]*a[i]%MOD)%MOD;
printf("%I64d\n",ans);
}
return 0;
}