2017 Multi-University Training Contest - Team 10 HDU-6172：Array Challenge

Array Challenge

                                                                Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 153428/153428
K (Java/Others)

                                                                                   Total Submission(s): 393    Accepted Submission(s): 206


Problem Description

There’s an array that is generated by following rule.
h0=2,h1=3,h2=6,hn=4hn−1+17hn−2−12hn−3−16

And let us define two arrays bnandan as
below.
bn=3hn+1hn+9hn+1hn−1+9h2n+27hnhn−1−18hn+1−126hn−81hn−1+192(n>0)
an=bn+4n

Now, you have to print ⌊√(an)⌋ ,
n>1.

Your answer could be very large so print the answer modular 1000000007.

 

Input

The first line of input contains T (1 <= T <= 1000) , the number of test cases.

Each test case contains one integer n (1 < n <= 1015)
in one line.

 

Output

For each test case print ⌊√(a_n )⌋ modular 1000000007.

 

Sample Input

3
4
7
9

 

Sample Output

1255
324725
13185773

官方题解：http://bestcoder.hdu.edu.cn/blog/



√(a_n )=h(n+1)+h(n)-6*h(n-1)+8；即矩阵快速幂；

#include<bits/stdc++.h>
using namespace std;
const long long MOD=1000000007;
struct lenka
{
long long a[4][4];
};
lenka cla(const lenka& a,const lenka& b)
{
lenka c;
memset(c.a,0,sizeof c.a);
for(int i=0;i<4;i++)
{
for(int j=0;j<4;j++)
{
for(int k=0;k<4;k++)
{
c.a[i][j]+=a.a[i][k]*b.a[k][j];
c.a[i][j]=(c.a[i][j]+MOD)%MOD;  //负数取模
}
}
}
return c;
}
long long POW(long long n)
{
if(n==3)return 31;
n-=4;
lenka res,a,ans;
memset(res.a,0,sizeof res.a);
memset(a.a,0,sizeof a.a);
memset(ans.a,0,sizeof ans.a);
for(int i=0;i<4;i++)res.a[i][i]=1;
a.a[0][0]=4,a.a[0][1]=1;
a.a[1][0]=17,a.a[1][2]=1;
a.a[2][0]=-12;
a.a[3][0]=-16,a.a[3][3]=1;
while(n)
{
if(n%2)res=cla(res,a);
a=cla(a,a);
n/=2;
}
ans.a[0][0]=190,ans.a[0][1]=35,ans.a[0][2]=6,ans.a[0][3]=1;//初始化为h4,h3,h2
ans=cla(ans,res);
return (((ans.a[0][0]+ans.a[0][1])%MOD-6*ans.a[0][2])%MOD+8ll+MOD)%MOD; //注意取模
}
int main()
{
int T;cin>>T;
while(T--)
{
long long n;
cin>>n;
cout<<POW(n+1)<<endl;
}
return 0;
}