HDU 1757 矩阵求第n的递推式

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1838    Accepted Submission(s): 1053

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

 

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 using namespace std;

 6 

 7 const int N= 13;

 8 

 9 struct node

10 {

11     __int64 mat[N][N];

12 }hxl;

13 int A[13],k,m;

14 

15 void make_first(node *cur)

16 {

17     int i,j;

18     for(i=1;i<=10;i++)

19     for(j=1;j<=10;j++)

20     if(i==j)

21     cur->mat[i][j]=1;

22     else cur->mat[i][j]=0;

23 }

24 

25 struct node cheng(node cur,node now)

26 {

27     node ww;

28     int i,j,k;

29     memset(ww.mat,0,sizeof(ww.mat));

30     for(i=1;i<=10;i++)

31     for(k=1;k<=10;k++)

32     if(cur.mat[i][k])

33     {

34         for(j=1;j<=10;j++)

35         if(now.mat[k][j])

36         {

37             ww.mat[i][j]+=cur.mat[i][k]*now.mat[k][j];

38             if(ww.mat[i][j]>=m)

39             ww.mat[i][j]%=m;

40         }

41     }

42     return ww;

43 }

44 

45 void power_sum2(int n)

46 {

47     __int64 sum=0;

48     int i;

49     node cur,now=hxl;

50     make_first(&cur);

51     while(n)

52     {

53         if(n&1)

54         {

55             cur=cheng(cur,now);

56         }

57         n=n>>1;

58         now=cheng(now,now);

59     }

60     for(i=1;i<=10;i++)

61     sum= (sum+(10-i)*cur.mat[1][i])%m;

62     printf("%I64d\n",sum);

63 

64 }

65 

66 void make_ini()

67 {

68     int i,j;

69     if(k<=9)

70     {

71         printf("%d\n",k);

72         return;

73     }

74     memset(hxl.mat,0,sizeof(hxl.mat));

75     for(i=1;i<=10;i++)

76     hxl.mat[1][i]=A[i];

77 

78     for(i=2;i<=10;i++)// 初始化

79     for(j=1;j<=10;j++)

80     if(i-j==1)

81     hxl.mat[i][j]=1;

82 

83     power_sum2(k-9);

84 }

85 

86 int main()

87 {

88     int i;

89     while(scanf("%d%d",&k,&m)>0)

90     {

91         for(i=1;i<=10;i++)

92         scanf("%d",&A[i]);

93         make_ini();

94       //  cs();

95     }

96     return 0;

97 }

 

 

Source

2007省赛集训队练习赛（6）_linle专场

 

 

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