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hdu 1757 A Simple Math Problem(矩阵快速幂)
2015-07-20 17:39:48 来源: 作者: 【 】 浏览:3
Tags:hdu 1757 Simple Math Problem 矩阵 快速

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2780 Accepted Submission(s): 1649


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
图解:
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这个图来自:http://www.cnblogs.com/wally/archive/2013/03/01/2938305.html

<??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHByZSBjbGFzcz0="brush:java;">#include"iostream" #include"stdio.h" #include"string.h" #include"algorithm" #include"queue" #include"vector" using namespace std; #define N 10 #define LL __int64 int M,f[N]; struct Mat { LL mat[N][N]; }; Mat operator *(Mat a,Mat b) { int i,j,k; Mat c; for(i=0;i >=1; a=a*a; } LL s=0; for(i=0;i


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