HDU 4291 A Short problem（找循环节+快速幂矩阵）

题目链接

在知道算法的情况下，写了矩阵的算法去找循环节，跑了10来分钟没出结果。。。真心2B啊。。。好好理解这种利用循环节的优化。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <cstdlib>

 6 using namespace std;

 7 #define LL __int64

 8 int len1 = 222222224,len2 = 183120;

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

10 LL qmod(LL n,LL MOD)

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 = 1; i <= 2; i ++)

20                 for(j = 1; j <= 2; j ++)

21                     for(k = 1; k <= 2; k ++)

22                     {

23                         c[i][j] = (c[i][j]+(mat[i][k]*p[k][j])%MOD)%MOD;

24                     }

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

26         }

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

28         for(i = 1; i <= 2; i ++)

29             for(j = 1; j <= 2; j ++)

30                 for(k = 1; k <= 2; k ++)

31                 {

32                     c[i][j] = (c[i][j]+(p[i][k]*p[k][j])%MOD)%MOD;

33                 }

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

35         n >>= 1;

36     }

37     return mat[1][1]%MOD;

38 }

39 int main()

40 {

41     int i,j,k;

42     LL n;

43 

44     while(scanf("%I64d",&n)!=EOF)

45     {

46         for(k = 1; k <= 3; k ++)

47         {

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

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

50             p[1][1] = 3;

51             p[1][2] = p[2][1] = 1;

52             p[2][2] = 0;

53             for(i = 1; i <= 2; i ++)

54                 for(j = 1; j <= 2; j ++)

55                     if(i == j)

56                         mat[i][j] = 1;

57                     else

58                         mat[i][j] = 0;

59             if(n == 0||n == 1)

60             {

61                 continue;

62             }

63             if(k == 1)

64             n = qmod(n-1,len2);

65             else if(k == 2)

66             n = qmod(n-1,len1);

67             else if(k == 3)

68             n = qmod(n-1,1000000007);

69         }

70         printf("%I64d\n",n);

71     }

72     return 0;

73 }

74 //找寻环节的程序

75 /*

76 #include <stdio.h>

77 int main()

78 {

79     LL i,j,k,t;

80     j = 1;k = 0;

81     for(i = 2;i <= 1000000000;i ++)

82     {

83         t = j;

84         j = (3*j + k)%222222224;

85         k = t;

86         if(j == 1&&k == 0)

87         {

88             printf("%d\n",i);

89             break;

90         }

91     }

92     return 0;

93 }

94 */