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Another kind of Fibonacci(hdu3306)

2024-08-15 13:50:26   325人阅读

Another kind of Fibonacci

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2353    Accepted Submission(s): 936


Problem Description
As we all known , the Fibonacci series : F(0) = 1, F(1) = 1, F(N) = F(N - 1) + F(N - 2) (N >= 2).Now we define another kind of Fibonacci : A(0) = 1 , A(1) = 1 , A(N) = X * A(N - 1) + Y * A(N - 2) (N >= 2).And we want to Calculate S(N) , S(N) = A(0)2 +A(1)2+……+A(n)2.

 

 

Input
There are several test cases.
Each test case will contain three integers , N, X , Y .
N : 2<= N <= 231 – 1
X : 2<= X <= 231– 1
Y : 2<= Y <= 231 – 1
 

 

Output
For each test case , output the answer of S(n).If the answer is too big , divide it by 10007 and give me the reminder.
 

 

Sample Input
2 1 1
3 2 3
 

 

Sample Output
6
196
 思路：矩阵快速幂；
S(n) = ∑f(n)2 = S(n-1)+f(n)2 = S(n-1)+x2f(n-1)2+y2f(n-2)2+2xyf(n-1)f(n-2)；
然后f(n)*f(n-1) = (x*f(n-1)+y*f(n-2))*f(n-1) = x*f(n-1)2+y*f(n-1)*f(n-2)；
然后构造矩阵；技术分享
其中的第三个矩阵写错了，应该是s[n-1];     f[n-1]^2;     f[n-2]^2;  f[n-1]*f[n-2];
 1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<string.h> 5 #include<queue> 6 #include<set> 7 #include<math.h> 8 #include<map> 9 using namespace std;10 typedef struct node11 {12         int m[4][4];13         node()14         {15                 memset(m,0,sizeof(m));16         }17 } maxtr;18 void Init(maxtr *ans,int x,int y);19 maxtr E();20 maxtr quick_m(maxtr ak,int m);21 const int mod = 10007;22 int main(void)23 {24         int n,x,y;25         while(scanf("%d %d %d",&n,&x,&y)!=EOF)26         {27                 int f1 = 2;28                 int a1 = 1;29                 int a0 = 1;30                 int xx = 1;31                 maxtr ask ;32                 Init(&ask,x,y);33                 maxtr tp = quick_m(ask,n-1);34                 printf("%d\n",(tp.m[0][0]*2+tp.m[0][1]*a1+tp.m[0][2]*a0+tp.m[0][3]*xx)%mod);35         }36         return 0;37 }38 void Init(maxtr *ans,int x,int y)39 {      memset(ans->m,0,sizeof(ans->m));40         x%=mod;y%=mod;41         ans->m[0][0] = 1;42         ans->m[0][1] = x*x%mod;43         ans->m[0][2] = y*y%mod;44         ans->m[0][3] =2*x*y%mod;45         ans->m[1][1] = x*x%mod;46         ans->m[1][2] = y*y%mod;47         ans->m[1][3] = 2*x*y%mod;48         ans->m[2][1] = 1;49         ans->m[3][1] = x%mod;50         ans->m[3][3] = y%mod;51 }52 maxtr E()53 {54         maxtr ak;55         int i,j;56         for(i = 0; i < 4; i++)57         {58                 for(j = 0; j < 4; j++)59                 {60                         if(i == j)61                                 ak.m[i][j]  = 1;62                 }63         }64         return ak;65 }66 maxtr quick_m(maxtr ak,int m)67 {68         int i,j;69         maxtr ac = E();70         while(m)71         {72                 if(m&1)73                 {74                         maxtr a;75                         for(i = 0; i < 4; i++)76                                 for(j = 0; j < 4; j++)77                                         for(int s= 0; s < 4; s++)78                                                 a.m[i][j] = (a.m[i][j] + ak.m[i][s]*ac.m[s][j]%mod)%mod;79                         ac = a;80                 }81                 maxtr b;82                 for(i = 0; i < 4; i++)83                         for(j = 0; j < 4; j++)84                                 for(int s = 0; s < 4; s++)85                                         b.m[i][j] = (b.m[i][j] + ak.m[i][s]*ak.m[s][j]%mod)%mod;86                 ak = b;87                 m>>=1;88         }89         return ac;90 }

 

Another kind of Fibonacci(hdu3306)


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