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

0

1

2

 

Sample Output

0

1

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：

 

#include<cstdio>
long long mo1=1000000007;
int main()
{
    long long i=1,a=1,b=0,c;
    while (i)
    {
        c=(3*a+b)%mo1;
        b=a;
        a=c;
        if (a==1&&b==0)
        {
            printf("%I64d
",i);
            break;
        }
        i++;
    }
}

ac代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long mo1=1000000007;
long long mo2=222222224;
long long mo3=183120;
long long s1[3][3],s2[3][3],s3[3][3];
long long f(long long n,long long mo)
{
    int i,j,k;
    s1[0][0]=s2[0][0]=3;
    s1[0][1]=s2[0][1]=1;
    s1[1][0]=s2[1][0]=1;
    s1[1][1]=s2[1][1]=0;
    n-=2;
    while (n)
    {
        if (n&1)
        {
            memset(s3,0,sizeof(s3));
            for (k=0;k<2;k++)
            for (i=0;i<2;i++){if (!s1[i][k]) continue;
            for (j=0;j<2;j++)
            s3[i][j]=(s3[i][j]+s1[i][k]*s2[k][j])%mo;}
            for (i=0;i<2;i++)
            for (j=0;j<2;j++) s2[i][j]=s3[i][j];
        }
        memset(s3,0,sizeof(s3));
        for (k=0;k<2;k++)
        for (i=0;i<2;i++) {if (!s1[i][k]) continue;
        for (j=0;j<2;j++)
        s3[i][j]=(s3[i][j]+s1[i][k]*s1[k][j])%mo;}
        for (i=0;i<2;i++)
        for (j=0;j<2;j++) s1[i][j]=s3[i][j];
        n>>=1;
    }
    return s2[0][0];
}
int main()
{
    long long n;
    long long ans;
    while (~scanf("%I64d",&n))
    {
        if (n==0) {printf("0
");continue;}
        if (n==1) {printf("1
");continue;}
        ans=f(n,mo3);
        if (ans>=2) //一定要有，不然ans<2时函数就会死循环。
        ans=f(ans,mo2);
        if (ans>=2)
        ans=f(ans,mo1);
        printf("%I64d
",ans);
    }
}