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

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 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 <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 # define INF 0x3f3f3f # 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 (. 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 ;}