HDU1757 A Simple Math Problem——————矩阵快速幂

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6573    Accepted Submission(s): 4033

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 ​​

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矩阵快速幂解决
第一步，构造矩阵：

然后利用矩阵快速幂求得：

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN =  13;

struct mat{ll m[MAXN][MAXN]; };

ll n,MOD;
ll N,k,b_n=0;
ll C[MAXN],h[MAXN];

mat mul(mat a,mat b)//矩阵乘法
{
        mat tmp;
        for(int i=1;i<=N;i++)
                for(int j=1;j<=N;j++)
                {
                        tmp.m[i][j]=0;
                        for(int k=1;k<=N;k++)
                                tmp.m[i][j]=(tmp.m[i][j]+a.m[i][k]*b.m[k][j]+MOD)%MOD;
                }
        return tmp;
}
mat pow_mod(mat a,ll n)
{
        mat ans;
        for(int i=1;i<=N;i++)//构造单位矩阵
                for(int j=1;j<=N;j++)
                        ans.m[i][j]=(i==j);
        while(n)//矩阵快速幂
        {
                if(n&1) ans=mul(ans,a);
                a=mul(a,a);
                n>>=1;
        }
        return ans;
}

void init(mat &res,mat &H)//构造矩阵
{
        for(int i=1;i<=N;i++)
                res.m[1][i]=C[i];
        res.m[1][N]=b_n;
        for(int i=2;i<=N;i++)
                for(int j=1;j<=N;j++)
                        res.m[i][j]=(i==j+1);
        res.m[N][N-1] = 0;
        res.m[N][N]   = 1;
        for(int i=1;i<=N;i++)
        {
                H.m[i][1]=h[N-i];
                for(int j=2;j<=N;j++)
                        H.m[i][j]=0;
        }
        H.m[N][1]=1;
}
void slove(ll k,ll n)
{
        mat res,H;
        init(res,H);
        res = pow_mod(res,n-k+1);
        res = mul(res,H);
        ll ans=res.m[1][1];

        printf("%lld\n",ans%MOD);
}

int main()
{
        k=10; N=11;
        for(int i=1;i<=k;i++)   h[i]=i-1;

        while(~scanf("%lld %lld",&n,&MOD))//f(n)%MOD;,我更习惯这样写
        {

                for(int i=1;i<=k;i++)
                        scanf("%lld",&C[i]);//第一次没有加&，找了好久。。。我好傻
                if(n<10)
                {
                        printf("%d\n",n);
                        continue;
                }
                slove(k,n);
        }
        return 0;
}