【HDU3802】【降幂大法+矩阵加速+特征方程】Ipad,IPhone

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*10 ⁹ , 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 }

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