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 <= 10 18), You should solve for
g(g(g(n))) mod 10
9 + 7
where
g(n) = 3g(n - 1) + g(n - 2)
g(1) = 1
g(0) = 0
Hence they prefer problems short, too. Here is a short one:
Given n (1 <= n <= 10 18), You should solve for
where
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).
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 }
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 }
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 }