Hdu1757 -- A Simple Math Problem (Rapid matrix Power Optimization)

Source: Internet
Author: User

Hdu1757 -- A Simple Math Problem (Rapid matrix Power Optimization)

A Simple Math Problem Time Limit:1000 MS Memory Limit:32768KB 64bit 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 cocould 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.bkjia.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 ;}

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