UVA 12470 - Tribonacci(快速幂矩阵）

题目链接

开始居然错以为是Fib，其实是Trib,对矩阵加深了一下认识，F(n) = F(n-1)+F(n-2)+F(n-3) 初始矩阵变为3阶了。幂模依旧很搓，搜很多个版本，基本上都是运算符重载的，还有调用数组名的，当然也有直接运算的，凑合着用吧。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <cstdlib>

 6 using namespace std;

 7 #define MOD 1000000009

 8 #define LL long long

 9 LL p[3][3],mat[3][3];

10 LL qmod(LL n)

11 {

12     int i,j,k;

13     LL c[3][3];

14     while(n)

15     {

16         if(n&1)

17         {

18             memset(c,0,sizeof(c));

19             for(i = 0; i <= 2; i ++)

20                 for(j = 0; j <= 2; j ++)

21                     for(k = 0; k <= 2; k ++)

22                     {

23                         c[i][j] += mat[i][k]*p[k][j];

24                         c[i][j] %= MOD;

25                     }

26             memcpy(mat,c,sizeof(mat));

27         }

28         memset(c,0,sizeof(c));

29         for(i = 0; i <= 2; i ++)

30             for(j = 0; j <= 2; j ++)

31                 for(k = 0; k <= 2; k ++)

32                 {

33                     c[i][j] += p[i][k]*p[k][j];

34                     c[i][j] %= MOD;

35                 }

36         memcpy(p,c,sizeof(p));

37         n >>= 1;

38     }

39     return (2*mat[0][0]+mat[0][1])%MOD;

40 }

41 int main()

42 {

43     int i,j;

44     LL n,ans;

45     while(scanf("%lld",&n)!=EOF)

46     {

47         if(n == 0)

48             break;

49         memset(p,0,sizeof(p));

50         p[0][1] = p[0][2] = p[0][0] = 1;

51         p[1][0] = 1;

52         p[2][1] = 1;

53         for(i = 0; i <= 2; i ++)

54             for(j = 0; j <= 2; j ++)

55                 if(i == j)

56                     mat[i][j] = 1;

57                 else

58                     mat[i][j] = 0;

59         if(n <= 3)

60         {

61             printf("%lld\n",n-1);

62             continue;

63         }

64         printf("%lld\n",qmod(n-3));

65     }

66     return 0;

67 }