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 a
0,1 = 233,a
0,2 = 2333,a
0,3 = 23333...) Besides, in 233 matrix, we got a
i,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?
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 ≤ 10 9). The second line contains n integers, a 1,0,a 2,0,...,a n,0(0 ≤ a i,0 < 2 31).
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).
Output
For 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 72937Hint
#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;
}