【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*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

Sample Output

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
", Ans);
110     }
111     //p = 0x7fffffff;
112     //printf("%lld", get(5));
113     //for (int i = 0; i <= 10; i++) printf("%lld
", get(i));
114     return 0;
115 }

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