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
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；
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 node 11 { 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 ",(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 }