Problem Description
In ACM_DIY, there is one master called “Lost”. As we know he is a “-2Dai”, which means he has a lot of money.
Well, Lost use Ipad and IPhone to reward the ones who solve the following problem.
In this problem, we define F( n ) as :
Then Lost denote a function G(a,b,n,p) as 
Here a, b, n, p are all positive integer!
If you could tell Lost the value of G(a,b,n,p) , then you will get one Ipad and one IPhone!
Input
The first line is one integer T indicates the number of the test cases. (T <= 100)
Then for every case, only one line containing 4 positive integers a, b, n and p.
(1 ≤a, b, n, p≤2*109 , p is an odd prime number and a,b < p.)
Output
Output one line,the value of the G(a,b,n,p) .
Sample Input
4
2 3 1 10007
2 3 2 10007
2 3 3 10007
2 3 4 10007
Sample Output
40
392
3880
9941
Author
AekdyCoin
Source
ACM-DIY Group Contest 2011 Spring
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【分析】
不知道怎么回事...跟网上的程序对拍了好像没错，怎么过不了...应该是有些比较坑的点....
不错的一道题目，前面的两项不说了，后面的那一项可以把二次方弄进去，然后变成一个用韦达定理做一个逆运算得到特征方程。
然后就有递推式了，然后就可以矩阵加速了。
然后因为幂实在是太大了，然后上降幂大法，然后没了。
其实还涉及到二次剩余的理论，如果前面没有那两个式子答案就错了....因为刚好是那两个式子，让不能用降幂大法的情况变成0...
1 /*
2 五代李煜
3 《相见欢·林花谢了春红》
4 林花谢了春红，太匆匆。无奈朝来寒雨晚来风。
5 胭脂泪，相留醉，几时重。自是人生长恨水长东。
6 */
7 #include <iostream>
8 #include <cstdio>
9 #include <ctime>
10 #include <cmath>
11 #include <algorithm>
12 #include <cstring>
13 #include <string>
14 #include <map>
15 #include <set>
16 #include <vector>
17 #define LOCAL
18 const int MAXN = 1000 + 10;
19 const int INF = 0x7fffffff;
20 using namespace std;
21 typedef long long ll;
22 ll mod;//代表取模的数字
23 ll check, a, b, n, p;
24 struct Matrix{
25 ll num[5][5];
26 //Matrix(){memset(num, 0, sizeof(num));}
27 };
28 //为了防止和第一种矩阵乘法搞混
29 Matrix Mod(Matrix A, Matrix B, ll k){
30 if (k == 0){//代表两种不同的乘法
31 Matrix c;
32 memset(c.num, 0, sizeof (c.num));
33 for (ll i = 1; i <= 2; i++)
34 for (ll j = 1; j <= 2; j++)
35 for (ll k = 1; k <= 2; k++){
36 ll tmp = (A.num[i][k] * B.num[k][j]);
37 if (check) tmp %= (p - 1);
38 c.num[i][j] += tmp;
39 if (check) c.num[i][j] %= (p - 1);
40 }
41 //一旦大于了p-1代表当前出现的斐波那契数列会大于phi(p)，可以使用降幂大法
42 if ((c.num[1][1] + c.num[1][2]) > (p - 1)) check = 1;
43 return c;
44 }else if (k == 1){
45 Matrix C;
46 memset(C.num, 0, sizeof(C.num));
47 for (ll i = 1; i <= 2; i++)
48 for (ll j = 1; j <= 2; j++)
49 for (ll k = 1; k <= 2; k++) {
50 C.num[i][j] += (A.num[i][k] * B.num[k][j]) % p;
51 C.num[i][j] = ((C.num[i][j] % p) + p) % p;
52 }
53 return C;
54 }
55 }
56 //得到第x位的斐波那契数，也就是获得指数
57 Matrix Matrix_pow(Matrix A, ll x, ll k){
58 if (x == 1) return A;
59 Matrix tmp = Matrix_pow(A, x / 2, k);
60 if (x % 2 == 0) return Mod(tmp, tmp, k);
61 else return Mod(Mod(tmp, tmp, k), A, k);
62 }
63 ll get(ll x){
64 if (x == 0) return 1;
65 else if (x == 1) return 1;
66 Matrix A, B;
67 A.num[1][1] = 1; A.num[1][2] = 1;
68 A.num[2][1] = 1; A.num[2][2] = 0;
69 x--;//为真实的需要乘的次数
70 if (x == 0) return 1;
71 B = Matrix_pow(A, x, 0);
72 if (B.num[1][1] + B.num[1][2] > (p - 1)) check = 1;
73 if (check == 0) return B.num[1][1] + B.num[1][2];
74 else return (B.num[1][1] + B.num[1][2]) % (p - 1) + p - 1;
75 }
76 //有了a,b,pos就可进行矩阵加速了
77 ll cal(ll a, ll b, ll pos){
78 if (pos == 0) return 2 % p;
79 else if (pos == 1) return (2 * (a + b)) % p;
80 Matrix A;
81 A.num[1][1] = (2 * (a + b)) % p; A.num[1][2] = (((-(a - b) * (a - b)) % p) + p) % p;
82 A.num[2][1] = 1; A.num[2][2] = 0;
83 pos--;
84 Matrix B;
85 B = Matrix_pow(A, pos, 1);
86 return (B.num[1][1] * A.num[1][1]) % p + (B.num[1][2] * 2) % p;
87 }
88 ll pow(ll a, ll b){
89 if (b == 0) return 1 % p;
90 if (b == 1) return a % p;
91 ll tmp = pow(a, b / 2);
92 if (b % 2 == 0) return (tmp * tmp) % p;
93 else return (((tmp * tmp) % p) * a) % p;
94 }
95
96 int main(){
97 int T;
98 scanf("%d", &T);
99 while (T--){
100 //for (int i = 1; i ，=)
101 scanf("%lld%lld%lld%lld", &a, &b, &n, &p);
102 check = 0;//判断f(n)是否大于p
103 ll pos = get(n);
104 ll Ans = cal(a, b, pos);
105 ll f1, f2;
106 f1 = (pow(a, (p - 1) / 2) + 1) % p;
107 f2 = (pow(b, (p - 1) / 2) + 1) % p;
108 Ans = (((f1 * f2) % p) * Ans) % p;
109 printf("%lld\n", Ans);
110 }
111 //p = 0x7fffffff;
112 //printf("%lld", get(5));
113 //for (int i = 0; i <= 10; i++) printf("%lld\n", get(i));
114 return 0;
115 }