hdu------(1757)A Simple Math Problem(简单矩阵快速幂)

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2791    Accepted Submission(s): 1659

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

 

 

Author

linle

 

 

Source

2007省赛集训队练习赛（6）_linle专场

 代码：

 1 //#define LOCAL

 2 #include<cstdio>

 3 #include<cstring>

 4 #define LL __int64

 5 using namespace std;

 6 const int maxn=10;

 7 LL k,m;

 8 int aa[maxn],mat[maxn][maxn];

 9 int ans[maxn][maxn];

10 

11 void init()

12 {

13     for(int i=0;i<10;i++)

14     {

15      for(int j=0;j<10;j++)

16      {

17        if(i==0)

18          mat[i][j]=aa[j];

19        else

20           if(i==j+1)mat[i][j]=1;

21        else mat[i][j]=0;

22        if(i==j)

23            ans[i][j]=1;

24        else ans[i][j]=0;

25      }

26     }

27 }

28 

29 void Matrix(int a[][10],int b[][10])

30 {

31 

32      int c[10][10];

33      for(int i=0;i<10;i++)

34      {

35          for(int j=0;j<10;j++)

36         {

37             c[i][j]=0;

38           for(int k=0;k<10;k++)

39            {

40             c[i][j]=(c[i][j]+a[i][k]*b[k][j])%m;

41           }

42         }

43      }

44     for(int i=0;i<10;i++)

45     {

46       for(int j=0;j<10;j++)

47       {

48         a[i][j]=c[i][j];

49       }

50     }

51 }

52 

53 void pow(LL n)

54 {

55     while(n>0)

56     {

57         if(n&1) Matrix(ans,mat);

58         n>>=1L;

59         if(n==0)break;

60         Matrix(mat,mat);

61     }

62 }

63 

64 int main()

65 {

66     #ifdef LOCAL

67      freopen("test.in","r",stdin);

68     #endif

69 

70   while(scanf("%I64d%I64d",&k,&m)!=EOF)

71   {

72        for(int i=0;i<10;i++)

73        scanf("%d",aa+i);

74      if(k<10) printf("%I64d\n",k%m);

75      else

76      {

77        init();

78        pow(k-9);

79        int res=0;

80        for(int i=0;i<10;i++)

81         res=(res+(10-i-1)*ans[0][i])%m;

82        printf("%d\n",res);

83     }

84   }

85   return 0;

86 }

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