hdu 4291 A Short problem (矩阵快速幂+循环节）

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

Sample Output

42837

这题主要找循环节，mo1=1000000007,mo2=222222224,mo3=183120;找到循环节后只要用矩阵快速幂就可以了，下面有找循环节代码。

由于找循环节很耗时，所以找到后直接写出来，不用写在代码里。

找循环节：

 1 #include<cstdio>
 2 long long mo1=1000000007;
 3 int main()
 4 {
 5     long long i=1,a=1,b=0,c;
 6     while (i)
 7     {
 8         c=(3*a+b)%mo1;
 9         b=a;
10         a=c;
11         if (a==1&&b==0)
12         {
13             printf("%I64d
",i);
14             break;
15         }
16         i++;
17     }
18 }

View Code

ac代码：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 long long mo1=1000000007;
 6 long long mo2=222222224;
 7 long long mo3=183120;
 8 long long s1[3][3],s2[3][3],s3[3][3];
 9 long long f(long long n,long long mo)
10 {
11     int i,j,k;
12     s1[0][0]=s2[0][0]=3;
13     s1[0][1]=s2[0][1]=1;
14     s1[1][0]=s2[1][0]=1;
15     s1[1][1]=s2[1][1]=0;
16     n-=2;
17     while (n)
18     {
19         if (n&1)
20         {
21             memset(s3,0,sizeof(s3));
22             for (k=0;k<2;k++)
23             for (i=0;i<2;i++){if (!s1[i][k]) continue;
24             for (j=0;j<2;j++)
25             s3[i][j]=(s3[i][j]+s1[i][k]*s2[k][j])%mo;}
26             for (i=0;i<2;i++)
27             for (j=0;j<2;j++) s2[i][j]=s3[i][j];
28         }
29         memset(s3,0,sizeof(s3));
30         for (k=0;k<2;k++)
31         for (i=0;i<2;i++) {if (!s1[i][k]) continue;
32         for (j=0;j<2;j++)
33         s3[i][j]=(s3[i][j]+s1[i][k]*s1[k][j])%mo;}
34         for (i=0;i<2;i++)
35         for (j=0;j<2;j++) s1[i][j]=s3[i][j];
36         n>>=1;
37     }
38     return s2[0][0];
39 }
40 int main()
41 {
42     long long n;
43     long long ans;
44     while (~scanf("%I64d",&n))
45     {
46         if (n==0) {printf("0
");continue;}
47         if (n==1) {printf("1
");continue;}
48         ans=f(n,mo3);
49         if (ans>=2) //一定要有，不然ans<2时函数就会死循环。
50         ans=f(ans,mo2);
51         if (ans>=2)
52         ans=f(ans,mo1);
53         printf("%I64d
",ans);
54     }
55 }