*HDU 1757 矩阵乘法

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4307    Accepted Submission(s): 2586

Problem Description

Lele now is thinking about a simple function f(x).

If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .

Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.

 

Input

The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.

 

Output

For each case, output f(k) % m in one line.

 

Sample Input

10 9999

1 1 1 1 1 1 1 1 1 1

20 500

1 0 1 0 1 0 1 0 1 0

 

Sample Output

45

104

 

Author

linle

 

Source

2007省赛集训队练习赛（6）_linle专场

 

题意：

If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .

上面的递推式计算f(k)。计算结果%m;

代码：

//普通递推会超时，用矩阵乘法。构造矩阵时右上角的(n-1)*(n-1)矩阵置为单位矩阵，第n行从 右 到 左 填入系数，快速幂之后第n行作为行矩阵乘递推式从 小 到 大 排列的列矩阵得到结果
//或者构造左下角的单位矩阵，第一行从 左 到 右 填入系数，快速幂之后第1行作为行矩阵乘递推式从 大 到 小 排列的列矩阵得到结果
#include<bitsstdc++.h>
using namespace std;
int a[12],f[12]={0,1,2,3,4,5,6,7,8,9};
int k,m;
struct Lu
{
    int mp[12][12];
}L1;
void init()
{
    memset(L1.mp,0,sizeof(L1.mp));
    for(int i=0;i<=8;i++)
    {
        L1.mp[i][i+1]=1;
    }
    for(int i=0;i<=9;i++)
    {
        L1.mp[9][i]=a[9-i];
    }
}
Lu mult(Lu a,Lu b)
{
    Lu c;
    for(int i=0;i<=9;i++)
    for(int j=0;j<=9;j++)
    {
        c.mp[i][j]=0;
        for(int k=0;k<=9;k++)
        c.mp[i][j]+=(a.mp[i][k]*b.mp[k][j])%m;
        c.mp[i][j]%=m;
    }
    return c;
}
Lu solve(int x)
{
    if(x==1)
    return L1;
    if(x&1)
    {
        Lu q=solve(x-1);
        return mult(q,L1);
    }
    else
    {
        Lu q=solve(x/2);
        return mult(q,q);
    }
}
int main()
{
    while(scanf("%d%d",&k,&m)!=EOF)
    {
        for(int i=0;i<=9;i++)
        scanf("%d",&a[i]);
        if(k<10)
        {
            printf("%d
",k%m);
            continue;
        }
        init();
        Lu tem=solve(k-9);
        int ans=0;
        for(int i=0;i<=9;i++)
        ans+=(tem.mp[9][i]*f[i])%m;
        printf("%d
",ans%m);
    }
    return 0;
}

//普通递推会超时，用矩阵乘法。
#include<bitsstdc++.h>
using namespace std;
int a[12],f[12]={9,8,7,6,5,4,3,2,1,0};
int k,m;
struct Lu
{
    int mp[12][12];
}L1;
void init()
{
    memset(L1.mp,0,sizeof(L1.mp));
    for(int i=1;i<=9;i++)
    {
        L1.mp[i][i-1]=1;
    }
    for(int i=0;i<=9;i++)
    {
        L1.mp[0][i]=a[i];
    }
}
Lu mult(Lu a,Lu b)
{
    Lu c;
    for(int i=0;i<=9;i++)
    for(int j=0;j<=9;j++)
    {
        c.mp[i][j]=0;
        for(int k=0;k<=9;k++)
        c.mp[i][j]+=(a.mp[i][k]*b.mp[k][j])%m;
        c.mp[i][j]%=m;
    }
    return c;
}
Lu solve(int x)
{
    if(x==1)
    return L1;
    if(x&1)
    {
        Lu q=solve(x-1);
        return mult(q,L1);
    }
    else
    {
        Lu q=solve(x/2);
        return mult(q,q);
    }
}
int main()
{
    while(scanf("%d%d",&k,&m)!=EOF)
    {
        for(int i=0;i<=9;i++)
        scanf("%d",&a[i]);
        if(k<10)
        {
            printf("%d
",k%m);
            continue;
        }
        init();
        Lu tem=solve(k-9);
        int ans=0;
        for(int i=0;i<=9;i++)
        ans+=(tem.mp[0][i]*f[i])%m;
        printf("%d
",ans%m);
    }
    return 0;
}