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hdu 5015 233 Matrix

2024-07-22 17:32:18   219人阅读

233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 392    Accepted Submission(s): 262


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
 

Source
2014 ACM/ICPC Asia Regional Xi‘an Online
 


题解及代码：

   

       矩阵快速幂的题目，推出矩阵就可以了，具体推的方法就是有矩阵的第一列推出第二列，以此类推就可以了。

       这里给出10的矩阵：

     




#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
const __int64 mod=10000007;
struct mat           //矩阵的定义
{
    __int64 t[13][13];
    void set()
    {
        memset(t,0,sizeof(t));
    }
} a,b,c;

mat multiple(mat a,mat b,int n,int p)  //矩阵相乘函数
{
    int i,j,k;
    mat temp;
    temp.set();
    for(i=0; i<n; i++)
        for(j=0; j<n; j++)
        {
            if(a.t[i][j]!=0)
                for(k=0; k<n; k++)
                    temp.t[i][k]=(temp.t[i][k]+a.t[i][j]*b.t[j][k]+p)%p;
        }
    return temp;
}

mat quick_mod(mat b,int n,int m,int p)
{
    mat t;
    t.set();
    for(int i=0;i<n;i++) t.t[i][i]=1;
    while(m)
    {
        if(m&1)
        {
            t=multiple(t,b,n,p);
        }
        m>>=1;
        b=multiple(b,b,n,p);
    }
    return t;
}

void init(int n)
{
  a.set();
  for(int i=0;i<=n;i++)
  {
    if(i==0)
    for(int j=0;j<=n;j++)
    a.t[j][i]=1;
    else if(i==1)
    for(int j=1;j<=n;j++)
    a.t[j][1]=10;
    else
    for(int j=i;j<=n;j++)
    a.t[j][i]=1;
  }
}


int main()
{
    int n,m;
    __int64 s[14];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        s[0]=3;
        s[1]=23;
        for(int i=2;i<=n+1;i++)
        {
            scanf("%I64d",&s[i]);
        }
        init(n+1);
        b=quick_mod(a,n+2,m,mod);
        __int64 ans=0;
        for(int i=0;i<=n+1;i++)
        ans=(ans+(s[i]*b.t[n+1][i])%mod)%mod;
        printf("%I64d\n",ans);
    }
    return 0;
}






hdu 5015 233 Matrix


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