hdu 4291 A Short problem (成都网络赛 矩阵乘法 递推式 求 循环节)

http://acm.hdu.edu.cn/showproblem.php?pid=4291

 题意：

A Short problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 739    Accepted Submission(s): 277

Problem Description

　　According to a research, VIM users tend to have shorter fingers, compared with Emacs users.
　　Hence they prefer problems short, too. Here is a short one:
　　Given n (1 <= n <= 10¹⁸), You should solve for
g(g(g(n))) mod 10⁹ + 7
　　where
g(n) = 3g(n - 1) + g(n - 2)
g(1) = 1
g(0) = 0

 

Input

　　There are several test cases. For each test case there is an integer n in a single line.
　　Please process until EOF (End Of File).

 

Output

　　For each test case, please print a single line with a integer, the corresponding answer to this case.

 

Sample Input

0 1 2

 

Sample Output

0 1 42837

 

题解：  首先 看到 它是 三层 求  g(n);  因为  每一层  g(g(n))  与  g(g(n)% 10^9 + 1)  不一定相同 ，所以 我们要 求出  内层的 循环节；

 在 最外层 为  10^9 + 7 的情况下 ，我们 可以 求出 次外 层的 循环节 是  222222224   在将  222222224 带入 可以求出  最内层的 循环节 ，为183120 ；

 1 //求循环节  
 2 #include <cstdio>  
 3 #include <set>  
 4 const __int64 M = 222222224;  
 5 using namespace std;  
 6 __int64 a[4], b[4];  
 7   
 8 int main()  
 9 {  
10     set<__int64> se;  
11     __int64 i;  
12     __int64 f1 = 1, f0 = 0;  
13     for(i = 2; ; ++ i)  
14     {  
15         __int64 f2 = (3 * f1 + f0) % M;  
16         if(f1 == 0 && f2 == 1) break;  
17         f0 = f1; f1 = f2;  
18     }  
19     printf("%I64d\n", i - 1);  
20     return 0;  
21 }

 在 矩阵乘法 时由于  将 成的 次数  忘了 变为  long long  的  错了  好几次 。。。。。。。。。。。。。。。。

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<cmath>
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<set>
  7 #include<map>
  8 #include<queue>
  9 #include<vector>
 10 #include<string>
 11 #define Min(a,b) a<b?a:b
 12 #define Max(a,b) a>b?a:b
 13 #define CL(a,num) memset(a,num,sizeof(a));
 14 #define eps  1e-12
 15 #define inf   10000000
 16 
 17 
 18 //freopen("data.txt","r",stdin);
 19 const double pi  = acos(-1.0);
 20 typedef   __int64  ll;
 21 const int maxn = 101000 ;
 22 using namespace std;
 23 struct   matrix
 24 {
 25     ll m[2][2];
 26 
 27 };
 28 int n = 2;
 29 ll mod ;
 30 matrix e;
 31 void init()
 32 {
 33    e.m[0][0] = 1;
 34    e.m[0][1] = 0;
 35    e.m[1][0] = 0;
 36    e.m[1][1] = 1;
 37 }
 38 
 39 matrix mtmul(matrix a,matrix b)
 40 {
 41 
 42     matrix c;
 43     int i,j,k;
 44     for(i = 0; i < n; i++)
 45     {
 46         for(j = 0; j < n;j++)
 47         {
 48             c.m[i][j] = 0;
 49             for(k = 0 ; k < n;k++)
 50             {
 51                 c.m[i][j] += a.m[i][k]* b.m[k][j];
 52                 c.m[i][j] %= mod;
 53             }
 54         }
 55     }
 56 
 57 
 58     return c;
 59 }
 60 matrix mtpow(matrix d, ll k)
 61 {   matrix a;
 62 
 63 
 64     a = e;//e 为单位矩阵
 65    while(k)
 66    {
 67        if(k&1)
 68        {
 69            a = mtmul(a,d);
 70        }
 71        k/=2;
 72        d = mtmul(d,d);
 73    }
 74    return a;
 75 
 76 }
 77 
 78 int main()
 79 {
 80     ll  num;
 81     init() ;
 82     //freopen("data.txt","r",stdin);
 83 
 84     //freopen("out.txt","w",stdout);
 85     while(cin>>num)
 86     {
 87 
 88         if(num == 0){printf("0\n");continue ;}
 89         if(num == 1){printf("1\n");continue ;}
 90 
 91           matrix a,c,d,b;
 92 
 93         a.m[0][0] =  3;
 94         a.m[0][1] = 1;
 95         a.m[1][0] = 1;
 96         a.m[1][1]  = 0 ;
 97 
 98         b.m[0][0] = 1;
 99         b.m[0][1] = 0;
100         b.m[1][0] = 0;
101         b.m[1][1] = 0;
102         mod = 183120 ;
103         c = mtpow(a,num - 1);
104        num = c.m[0][0] ;
105 
106 
107         if(c.m[0][0] == 0){printf("0\n");continue ;}
108         if(c.m[0][0] == 1){printf("1\n");continue ;}
109 
110 
111 
112 
113         mod = 222222224 ;
114         c = mtpow(a,num - 1);
115 
116         num = c.m[0][0] ;
117 
118 
119          if(c.m[0][0] == 0){printf("0\n");continue ;}
120         if(c.m[0][0] == 1){printf("1\n");continue ;}
121 
122 
123 
124 
125         mod = 1000000007 ;
126         c = mtpow(a,num - 1);
127 
128 
129         num = c.m[0][0] ;
130          if(c.m[0][0] == 0){printf("0\n");continue ;}
131         if(c.m[0][0] == 1){printf("1\n");continue ;}
132 
133 
134        cout<<num<<endl;
135 
136     }
137 }