【矩阵快速幂】【杭电OJ1757】

http://acm.hdu.edu.cn/showproblem.php?pid=1757

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5829    Accepted Submission(s): 3555

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

题目分析：看到 1）f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);【一个递推式】

　　　　　　　　2） k<2*10^9 , m < 10^5【数据超大】

　　　　　　就知道是矩阵快速幂了【好吧..我之前并不知道..qaq..】

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 using namespace std;
 long long n,k;
 void mul(long long wqw[],long long tat[][])
 {
     long long or2[];
 //    cout <<
     memset(or2,,sizeof(or2));
     for(int i =  ; i <   ; i++)
     {
         for(int j =  ; j <  ; j++)
         {
             or2[i]+=(wqw[j]*(tat[j][i]))%k;
             or2[i]%=k;
         }
     }
     memcpy(wqw,or2,sizeof(or2));
 }
 void mulself(long long awa[][])
 {
     long long or3[][];
     memset(or3,,sizeof(or3));
     for(int i = ; i <  ; i++)
     {
         for(int j =  ; j<  ; j++)
         {
             for(int kk =  ; kk <  ; kk++)
             {
                 or3[i][j]+=((awa[i][kk])%k)*((awa[kk][j])%k)%k;
                 or3[i][j]%=k;
             }
         }
     }
     memcpy(awa,or3,sizeof(or3));
 }
 int main()
 {
     while(scanf("%lld%lld",&n,&k)==)
     {
 
     long long qaq[][];
     long long qwq[];
     long long orz[]={,,,,,,,,,};
     memset(qaq,,sizeof(qaq));
     for(int i =  ; i <  ; i++)
     {
         scanf("%lld",&qaq[i][]);
     }
     for(int i =  ; i < ; i++)
     {
         qaq[i-][i]=;
     }
     if(n<)
     {
         printf("%lld\n",n%k);
     }
     else
     {
         n=n-;
         while(n)
         {
             if(n&)mul(orz,qaq);
             mulself(qaq);
             n/=;
         }
         cout  <<orz[]%k<<endl;
     }
 }
     return ;
 }