hdu1757 A Simple Math Problem（矩阵快速幂优化） -电脑资料

    A Simple Math ProblemTime Limit:1000MSMemory Limit:32768KB64bit IO Format:%I64d & %I64uSubmit Status

    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 99991 1 1 1 1 1 1 1 1 120 5001 0 1 0 1 0 1 0 1 0

    Sample Output

45104

   

    <喎?http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+PC9wPgo8cHJlIGNsYXNzPQ=="brush:java;">#include#include#include using namespace std ;#define LL __int64struct node{ LL a[12][12] ; int n ;};node mul(node p,node q,LL m){ int i , j , k ; node s ; s.n = p.n ; for(i = 0 ; i< p.n ; i++) for(j = 0 ; j< p.n ; j++) { s.a[i][j] = 0 ; for(k = 0 ; k< p.n ; k++) s.a[i][j] = (s.a[i][j] + p.a[i][k]*q.a[k][j]) % m ; } return s ;}node pow(node p,LL k,LL m){ if( k == 1 ) return p ; node s = pow(p,k/2,m) ; s = mul(s,s,m) ; if( k%2 ) { s = mul(s,p,m) ; } return s ;}int main(){ LL k , m , ans ; int i , j ; node p , s ; while( scanf("%I64d %I64d", &k, &m) != EOF ) { if(k< 10) { printf("%I64d\n", k) ; continue ; } ans = 0 ; p.n = 10 ; for(i = 0 ; i< 10 ; i++) { for(j = 0 ; j< 10 ; j++) p.a[i][j] = 0 ; p.a[i][i+1] = 1 ; } for(i = 0 ; i< 10 ; i++) scanf("%I64d", &p.a[i][0]) ; s = pow(p,k-9,m) ; for(i = 0 ; i< 10 ; i++) ans = ( ans + (9-i)*s.a[i][0] ) % m ; printf("%I64d\n", ans) ; } return 0 ;}