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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posted @ 2013-08-18 12:06  芷水  阅读(197)  评论(0)  编辑  收藏  举报