[Hdu 4959] Poor Akagi number theory (Lucas number, quadratic field operation, proportional sequence sum)

[Hdu 4959] Poor Akagi number theory (Lucas number, quadratic field operation, proportional sequence sum)
Poor AkagiTime Limit: 30000/15000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 131 Accepted Submission (s): 29


Problem DescriptionAkagi 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.
InputThis 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.
OutputFor each test case, you need to output the answer mod 1000000007 in a line.
Sample Input

33 12 24 3

Sample Output

1014443

SourceBestCoder Round #5
Theme
Calculate the first n items of the k power of the Lucas number and the number of Lucas is L [0] = 2, L [1] = 1, L [n] = L [N-2] + L [n-1] (n> = 2)
At the time, I thought that I could find the second surplus based on zoj 3774 ...... The result shows that 1e9 + 7 does not have a quadratic surplus. Finally, a clever method is found, which is based on the general formula of the Lucas number.
Set the sum formula


Define a quadratic field

In this case, the quadratic field can be directly added and multiplied (the final result is an integer, so root number 5 will not exist in the result)
Reload the plus and multiplication signs of the quadratic field, and define the Fast Power Operation of the quadratic field. Just enter the formula. =. = It seems that the data of hangdian in this question has not been corrected. For a moment, the public ratio is directly returned n + 1 (may cause overflow). Even if it is AC, then the positive solution is still WA ......
Here, only the code is corrected.

/***** Author: ahm001 ***** source: hdu 4959 ***** time: 08/21/14 ***** type: math *****/# include
 
  
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               # Define sqr (x) * (x) # define LL 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 (. first + B. first) % mod, (. second + B. second) % mod);} qf operator * (qf a, qf B) {// if (LL). first * (LL) B. first) % mod + (LL). second * (LL) B. second) % mod * 5ll) % mod <0) // printf ("% d \ n",. first,. second, B. first, B. second); if (. first <0). first + = mod; if (B. first <0) B. first + = mod; if (. second <0). second + = mod; if (B. second <0) B. second + = mod; return make_pair (LL). first * (LL) B. first) % mod + (LL). second * (LL) B. second) % mod * 5ll) % mod, (LL). first * (LL) B. second) % mod + (LL). second * (LL) B. first) % 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 +); // after this statement is removed, the AC is removed. However, if n + 1 is not moduled, qf powe = a will be cracked in the subsequent results; 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 (100000); r2 [0] = make_pair (); for (int I = 1; I <=; 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 \ n", ans. first, ans. second); // printf ("% d \ n", cnt);} return 0 ;}