[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   Recommend notonlysuccess   |   We have carefully selected several similar problems for you:   3805  3800  3801  3803  3804 [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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转载于:https://www.cnblogs.com/hoskey/p/4354052.html

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