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

Sample Output

10
14
443

Source BestCoder Round #5
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根据 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很巧妙的做法直接根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、乘操作即可(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，定义二次域的快速幂运算，全部带入公式即可。 =.=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）竟然AC了然后正解依然WA……
这里只放正解代码

/**
**author : ahm001  **
**source : hdu 4959**
**time :  08/21/14 **
**type : math      **
**/
#include 
  
   
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              #include 
             
               #include 
               #include 
               
                 #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 
                
                  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，但是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; }


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