Poor Akagi
Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 131 Accepted Submission(s): 29
Problem Description
Akagi is not only good at basketball but also good at math. Recently, he got a sequence Ln from his teacher. Ln is defined as follow:
$$\Large L(n)=\begin{cases}
2 & \text{ if } n=0 \\
1 & \text{ if } n=1 \\
L(n-1)+L(n-2) & \text{ if } n>1
\end{cases}$$
And Akagi’s teacher cherishes Agaki’s talent in mathematic. So he wants Agaki to spend more time studying math rather than playing basketball. So he decided to ask Agaki to solve a problem about Ln and promised that as soon as he solves this problem, he can
go to play basketball. And this problem is:
Given N and K, you need to find \(\Large\sum\limits_{0}^{N}L_i^K\)
And Agaki needs your help.
Input
This problem contains multiple tests.
In the first line there’s one number T (1 ≤ T ≤ 20) which tells the total number of test cases. For each test case, there an integer N (0 ≤ N ≤ 10^18) and an integer K (1 ≤ K ≤ 100000) in a line.
Output
For each test case, you need to output the answer mod 1000000007 in a line.
Sample Input
3 3 1 2 2 4 3
Sample Output
10 14 443
Source
题目大意
求卢卡斯数的k次方的前n项和
卢卡斯数为L[0]=2,L[1]=1,L[n]=L[n-2]+L[n-1](n>=2)
题目思路
当时看到题还以为直接依据 zoj 3774 找出二次剩余…… 结果发现1e9+7不存在二次剩余
最后发现了一种非常巧妙的做法
直接依据卢卡斯数的通项公式
设
则求和公式为
定义二次域
此时直接对二次域进行加、乘操作就可以(最后的结果为整数，故根号五不会存在在结果之中)
重载二次域的加号和乘号，定义二次域的高速幂运算。所有带入公式就可以。
=.=好像这一题的杭电的数据还没有修正
公比为一时直接返回n+1（可能带来溢出）居然AC了
然后正解依旧WA……
这里仅仅放正解代码
/**
**author : ahm001 **
**source : hdu 4959**
**time : 08/21/14 **
**type : math **
**/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#define sqr(x) ((x)*(x))
#define LL long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define eps 1e-10
#define mod 1000000007
using namespace std;
int cnt=0;
typedef pair <LL,LL> qf;
qf operator + (qf a,qf b)
{
return make_pair((a.first+b.first)%mod,(a.second+b.second)%mod);
}
qf operator * (qf a,qf b)
{
// if ((((LL)a.first*(LL)b.first)%mod+((LL)a.second*(LL)b.second)%mod*5ll)%mod<0)
// printf("%d %d %d %d\n",a.first,a.second,b.first,b.second);
if (a.first<0) a.first+=mod;
if (b.first<0) b.first+=mod;
if (a.second<0) a.second+=mod;
if (b.second<0) b.second+=mod;
return make_pair((((LL)a.first*(LL)b.first)%mod+((LL)a.second*(LL)b.second)%mod*5ll)%mod,
(((LL)a.first*(LL)b.second)%mod+((LL)a.second*(LL)b.first)%mod)%mod);
}
qf pow(qf a, LL x)
{
qf res(1,0);
qf multi=a;
while (x)
{
if (x&1)
{
res=res*multi;
}
multi=multi*multi;
x/=2;
}
return res;
}
LL pow(LL a,LL b)
{
LL res=1;
LL multi=a;
while (b)
{
if (b&1)
{
res=res*multi%mod;
}
multi=multi*multi%mod;
b/=2;
}
return res;
}
qf acce(qf a,LL b)
{
qf ans=make_pair(1,0);
// if (a==ans) return make_pair(b+1,0);//这条语句去掉后AC了。可是n+1不取模将会造成后面的结果爆掉
qf powe=a;
qf sum=a;
qf multi=make_pair(1,0);
while (b)
{
if (b&1)
{
ans=ans+(multi*sum);
multi=multi*powe;
}
sum=sum*(powe+make_pair(1,0));
powe=powe*powe;
b/=2;
}
return ans;
}
LL inv[100005];
qf r1[100005],r2[100005];
void egcd (LL a,LL b,LL &x,LL &y)
{
if (b==0)
{
x=1,y=0;
return ;
}
egcd(b,a%b,x,y);
LL t=x;
x=y;y=t-a/b*y;
}
int main()
{
LL x,y;
for (LL i=1;i<=100000;i++)
{
egcd(i,mod,x,y);
x=(x+mod)%mod;
inv[i]=x;
}
r1[0]=make_pair(1,0);
r2[0]=make_pair(1,0);
for (int i=1;i<=100000;i++)
{
r1[i]=r1[i-1]*make_pair(1,1);
r2[i]=r2[i-1]*make_pair(1,-1);
}
int T;
scanf("%d",&T);
while (T--)
{
cnt=0;
LL n,m;
scanf("%I64d%I64d",&n,&m);
// n=1e18;
// m=1e5;
qf ans=make_pair(0,0);
LL Ca=1;
LL v=pow(inv[2],m);
for (LL i=0;i<=m;i++)
{
// printf("%lld\n",Ca);
qf p(Ca,0);
qf tmp=r1[i]*r2[m-i]*make_pair(v,0);
tmp=acce(tmp,n);
tmp=tmp*p;
ans=ans+tmp;
Ca=Ca*(m-i)%mod;
Ca=Ca*inv[i+1]%mod;
}
LL aa=(LL)ans.first;
printf("%I64d\n",aa);
// printf("%d %d \n",ans.first,ans.second);
// printf("%d\n",cnt);
}
return 0;
}