8.3.2 A Simple Math Problem

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 44 Accepted Submission(s): 42

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

思路：遞推數列均可以用矩陣解的～例如一個遞推數列Fn=Fn-1+Fn-2+..+Fn-k

構造Fk,Fk+1,Fk+2..Fn-1,Fn

 --->Fk+1,Fk+2,....Fn,Fn+1便可……

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <string>
 6 #include <cstdlib>
 7 using namespace std;
 8 
 9 typedef long long ll;
10 const int maxn=10;
11 struct matrix
12 {
13     int m[maxn][maxn];
14 };
15 matrix c,base,ans;
16 int n,k,mod,sum,x;
17 int a[maxn];
18 
19 
20 void close()
21 {
22 exit(0);
23 }
24 
25 matrix mul(matrix a,matrix b)
26 {
27     memset(c.m,0,sizeof(c.m));
28     for (int i=0;i<=9;i++)
29         for (int j=0;j<=9;j++)
30             for (int k=0;k<=9;k++)
31                 c.m[i][j]=(c.m[i][j]+(ll)a.m[i][k]*b.m[k][j]) % mod;
32     return c;
33 }
34 
35 void print(matrix a)
36 {
37     for (int i=0;i<=9;i++)
38     {
39         for (int j=0;j<=9;j++)
40             printf("%d ",a.m[i][j]);
41         printf("\n");
42     }
43 }
44 
45 void work()
46 {
47     k-=9;
48     memset(ans.m,0,sizeof(ans.m));
49     for (int i=0;i<=9;i++)
50         ans.m[i][i]=1;
51     while (k!=0)
52     {
53         if (k & 1)
54             ans=mul(base,ans);
55         k/=2;
56         base=mul(base,base);
57     }
58     sum=0;
59     for (int i=0;i<=9;i++)
60         sum=(sum + (ll)(i)*ans.m[9][i]) % mod;
61     printf("%d\n",sum);
62 }
63 
64 
65 void init()
66 {
67     while(scanf("%d %d",&k,&mod)!=EOF)
68     {
69         if (k<=9)
70         {
71             printf("%d\n",k);
72             continue;
73         }
74         for (int i=0;i<=9;i++)
75             scanf("%d",&a[9-i]);
76         memset(base.m,0,sizeof(base.m));
77         for (int i=0;i<=8;i++)
78             base.m[i][i+1]=1;
79         for (int i=0;i<=9;i++)
80             base.m[9][i]=a[i];
81         work();
82     }
83 }
84 
85 int main ()
86 {
87     init();
88     close();
89     return 0;
90 }