【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[10],long long tat[10][10])
{
    long long or2[10];
//    cout << 
    memset(or2,0,sizeof(or2));
    for(int i = 0 ; i < 10  ; i++)
    {
        for(int j = 0 ; j < 10 ; j++)
        {
            or2[i]+=(wqw[j]*(tat[j][i]))%k;
            or2[i]%=k;
        }
    }
    memcpy(wqw,or2,sizeof(or2));
}
void mulself(long long awa[10][10])
{
    long long or3[10][10];
    memset(or3,0,sizeof(or3));
    for(int i = 0; i < 10 ; i++)
    {
        for(int j = 0 ; j< 10 ; j++)
        {
            for(int kk = 0 ; kk < 10 ; 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)==2)
    {
    
    long long qaq[10][10];
    long long qwq[10];
    long long orz[10]={9,8,7,6,5,4,3,2,1,0};
    memset(qaq,0,sizeof(qaq));
    for(int i = 0 ; i < 10 ; i++)
    {
        scanf("%lld",&qaq[i][0]);
    }
    for(int i = 1 ; i < 10; i++)
    {
        qaq[i-1][i]=1;
    }
    if(n<10)
    {
        printf("%lld\n",n%k);
    }
    else
    {
        n=n-9;
        while(n)
        {
            if(n&1)mul(orz,qaq);
            mulself(qaq);
            n/=2;
        }
        cout  <<orz[0]%k<<endl;
    }
}
    return 0;
}