Let us define a sequence as below
⎧⎩⎨⎪⎪⎪⎪⎪⎪F1F2Fn===ABC⋅Fn−2+D⋅Fn−1+⌊Pn⌋{F1=AF2=BFn=C⋅Fn−2+D⋅Fn−1+⌊Pn⌋
Your job is simple, for each task, you should output FnFn module 109+7109+7.
Input
The first line has only one integer TT, indicates the number of tasks.
Then, for the next TT lines, each line consists of 66 integers, AA , BB, CC, DD, PP, nn.
1≤T≤200≤A,B,C,D≤1091≤P,n≤1091≤T≤200≤A,B,C,D≤1091≤P,n≤109
Output
36 24
Sample Input
2 3 3 2 1 3 5 3 2 2 2 1 4
Sample Output
36 24
思路:因为p/n向下取整的不同值个数会在p的根号左右,所以可以根据不同的p/i值进行分段做矩阵快速幂。
注意写代码时讨论n,p的大小关系就行。
代码:
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
using namespace std;
const long long mod=1e9+7;
struct jz
{
long long num[3][3];
jz() { memset(num,0,sizeof(num)); }
jz operator*(const jz &P)const
{
jz ans;
for(int k=0;k<3;k++)
for(int i=0;i<3;i++)
for(int j=0;j<3;j++)
ans.num[i][j]=(ans.num[i][j]+num[i][k]*P.num[k][j]%mod)%mod;
return ans;
}
}COE,ans,unit;
int T_T;
long long A,B,C,D,P,n;
jz pOw(jz X,long long m)
{
jz ans;
for(ans=unit;m;m>>=1,X=X*X)
if(m&1)
ans=ans*X;
return ans;
}
void init(long long A,long long B,long long C,long long D,long long x)
{
COE.num[0][0]=0;
COE.num[0][1]=1;
COE.num[0][2]=0;
COE.num[1][0]=C;
COE.num[1][1]=D;
COE.num[1][2]=x; // this element need to be changed each step.
COE.num[2][0]=0;
COE.num[2][1]=0;
COE.num[2][2]=1;
return;
}
int main()
{
for(int i=0;i<3;i++) unit.num[i][i]=1;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%lld%lld%lld%lld%lld%lld",&A,&B,&C,&D,&P,&n);
if(n==1) printf("%lld\n",A);
else if(n<P)
{
ans.num[0][0]=A;
ans.num[1][0]=B;
ans.num[2][0]=1;
for(long long i=3;i<=n;i=P/(P/i)+1)
{
init(A,B,C,D,P/i);
if(n<=P/(P/i)) COE=pOw(COE,n-i+1);
else COE=pOw(COE,P/(P/i)+1-i);
ans=COE*ans;
}
printf("%lld\n",ans.num[1][0]);
}
else if(P<=n)
{
ans.num[0][0]=A;
ans.num[1][0]=B;
ans.num[2][0]=1;
for(long long i=3;i<=P;i=P/(P/i)+1)
{
init(A,B,C,D,P/i);
COE=pOw(COE,P/(P/i)+1-i);
ans=COE*ans;
}
init(A,B,C,D,0);
COE.num[1][2]=0;
COE=pOw(COE,n-max(P,2LL));
ans=COE*ans;
printf("%lld\n",ans.num[1][0]);
}
}
return 0;
}