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HDU 1757　A Simple Math Problem（矩阵快速幂）

2024-11-20 04:47:02   203人阅读

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2930    Accepted Submission(s): 1762


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
104ＣＯＤＥ：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define N 100010
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos ( -1.0 );
const double E = 2.718281828;
typedef long long ll;

const int INF = 1000010;

using namespace std;

typedef vector<int>vec;
typedef vector<vec>mat;
int m,fac[11];
ll k;

void init()
{
    for(int i=0;i<10;i++)
        fac[i]=i;
}

mat mul(mat &A,mat &B,int Mod)
{
    mat C(A.size(),vec(B[0].size()));
    for(int i=0; i<A.size(); i++)
    {
        for(int k=0; k<B.size(); k++)
        {
            for(int j=0; j<B[0].size(); j++)
            {
                C[i][j]=(C[i][j]+A[i][k]*B[k][j])%Mod;
            }
        }
    }
    return C;
}

mat pow(mat A,ll n,int Mod)
{
    mat B(A.size(),vec(A.size()));
    for(int  i=0; i<A.size(); i++)
    {
        B[i][i]=1;
    }
    while(n>0)
    {
        if(n&1)
            B=mul(B,A,Mod);
        A=mul(A,A,Mod);
        n>>=1;
    }
    return B;
}

int main()
{
    init();
    while(cin>>k>>m)
    {
        if(k<10)
        {
            printf("%d\n",fac[k]%m);
            continue;
        }
        mat A(10,vec(10));
        for(int i=0; i<10; i++)
        {
            scanf("%d",&A[0][i]);
        }
        for(int i=1;i<10;i++)
        {
            for(int j=0;j<10;j++)
            {
                if(i==j+1)
                    A[i][j]=1;
                else
                    A[i][j]=0;
            }
        }
        A=pow(A,k-9,m);
        int ans=0;
        for(int i=0;i<10;i++)
        {
            ans+=A[0][i]*fac[9-i];
            if(ans>=m)
                ans%=m;
        }
        cout<<ans<<endl;
    }
    return 0;
}


HDU 1757　A Simple Math Problem（矩阵快速幂）


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