2014西安網路賽1009||hdu5015 矩陣

life4711發表於2014-09-15
矩陣

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





解題思路：這題關鍵是矩陣的構造，比賽時是隊友推出的公式，程式碼實現如下，細節讀者品味，我就不羅嗦了。。。






#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;
}





            




相關文章