POJ 2118 矩阵乘法

题意：

_an=Σ_{1<=i<=k （}a_n-i*b_i）mod 10000

分析：

典型的矩阵乘法解线性递推式~

无限YM MATRIX67神犇~

http://www.matrix67.com/blog/archives/276/comment-page-1#comment-223435

View Code

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <iostream>
 5 
 6 #define SIZE 110
 7 #define mod 10000
 8 
 9 using namespace std;
10 
11 int n,m,a[SIZE],b[SIZE];
12 
13 struct MT
14 {
15     int x,y;
16     int mt[SIZE][SIZE];
17 }ans,def;
18 
19 inline MT operator *(MT a,MT b)
20 {
21     MT c;
22     memset(c.mt,0,sizeof c.mt);
23     c.x=a.x; c.y=b.y;
24     for(int i=1;i<=a.x;i++)
25         for(int j=1;j<=b.y;j++)
26             for(int k=1;k<=b.x;k++)
27                 c.mt[i][j]=(c.mt[i][j]+(a.mt[i][k]%mod)*(b.mt[k][j]%mod))%mod;
28     return c;
29 }
30 
31 bool read()
32 {
33     for(int i=0;i<n;i++) scanf("%d",&a[i]);
34     for(int i=0;i<n;i++) scanf("%d",&b[i]);
35     scanf("%d",&m);
36     if(m<=n-1) {printf("%d\n",a[m]);return false;}
37     m=m-(n-1);
38     for(int i=1;i<=n;i++) ans.mt[i][1]=a[i-1];
39     ans.x=n; ans.y=1;
40     memset(def.mt,0,sizeof def.mt);
41     for(int i=1;i<n;i++) def.mt[i][i+1]=1;
42     for(int i=1;i<=n;i++) def.mt[n][i]=b[n-i];
43     def.x=def.y=n;
44     return true;
45 }
46 
47 void go()
48 {
49     while(m)
50     {
51         if(m&1) ans=def*ans;
52         def=def*def;
53         m>>=1;
54     }
55     printf("%d\n",ans.mt[n][1]);
56 }
57 
58 int main()
59 {
60     while(scanf("%d",&n),n)
61     {
62         if(!read()) continue;
63         else go();
64     }
65     return 0;
66 }

没有人能阻止我前进的步伐，除了我自己！