• HDU 1757 A Simple Math Problem


    A Simple Math Problem

    Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 108 Accepted Submission(s): 77
     
    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
     

    典型的矩阵快速幂递推。

    很典型地构造一个

    f(n-1)    f(n-2)    f( n-3)  ....... f(n-10 )                       a0   1   0  0  0 ..........0  0 

     0             0             0      ........    0                            a1   0   1  0  0 ..........0  0 

    ......................................................              X           .......................................

     ....................................................                            a8  0  0  0 ...............0   1       

    0             0             0      ........    0                             a9   0  0  0 ..............0   0

    这样的矩阵来递推输出 最后一个矩阵的 ( 0 , 0 )位置的数就OK了~

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    using namespace std;
    typedef long long LL;
    const int N = 12 ;
    int n , mod , x[N];
    
    struct Mat
    {
        int c[N][N];
        Mat(){ memset( c , 0 , sizeof c ); };
        Mat operator * (const Mat &a )const{
            Mat res ;
            for( int i = 0 ; i < 10 ; ++i ){
                for( int k = 0 ; k < 10 ; ++k ){
                    if( !c[i][k] ) continue ;
                    for( int j = 0 ; j < 10 ; ++j ){
                        res.c[i][j] += c[i][k] * a.c[k][j] ;
                        res.c[i][j] %= mod;
                    }
                }
            }
            return res ;
        }
        void print(){
            for( int i =0 ; i < 10 ;++i ){
                for( int j = 0 ; j < 10 ;++j ){
                    cout << c[i][j] <<' ';
                }
                cout<<endl;
            }
        }
    };
    
    Mat quick_mod( Mat a , LL n )
    {
        Mat res = a ;
        while(n)
        {
            if( n & 1 )  res = res * a;
            a = a * a ;
            n >>= 1 ;
        }
        return res;
    }
    
    int main()
    {
        #ifdef LOCAL
            freopen("in.txt","r",stdin);
        #endif // LOCAL
        ios::sync_with_stdio(false);
        while( cin >> n >> mod ){
            for( int i = 0; i < 10 ; ++i ){
                cin >> x[i] ;
            }
            if( n < 10 ){
                cout << ( n % mod ) <<endl;
                continue ;
            }
            Mat a ,res ;
            for( int i = 0 ; i < 10 ; ++i )
                res.c[0][i] = ( 9 - i );
            for( int i = 0 ; i < 10 ; ++i ){
                a.c[i][0] = x[i] ;
                if( i < 9 ) a.c[i][i+1] = 1;
            }
    //        res.print();cout<<endl;
    //        a.print();cout<<endl;
    //        a = res * a;
    //        a.print();cout<<endl;
            res = res * quick_mod( a , n - 10 ) ;
    //        res.print();cout<<endl;
            cout << res.c[0][0] <<endl;
        }
    }
    only strive for your goal , can you make your dream come true ?
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  • 原文地址:https://www.cnblogs.com/hlmark/p/4001220.html
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