Another kind of Fibonacci
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
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 <= 2 31 ? 1
X : 2<= X <= 2 31? 1
Y : 2<= Y <= 2 31 ? 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
Sn = S(n-1) + An^2 = S(n-1) + X^2A*(n-1)^2 + 2*X*Y*A(n-1)*A(n-2) + Y^2*A(n-2)^2
也就得到Sn由S(n-1) , A(n-1) , A(n-2) , A(n-1)*A(n-2)得到,由此可以转化为矩阵相乘
#include <cstdio> #include <cstring> #include <algorithm> using namespace std ; #define LL __int64 #define MOD 10007 struct node { LL a[5][5] ; int n ; }; node mul(node p,node q) { int i , j , k ; node s ; for(i = 0 , s.n = p.n ; i < p.n ; i++) for(j = 0 ; j < p.n ; j++) { s.a[i][j] = 0 ; for(k = 0 ; k < p.n ; k++) s.a[i][j] = ( s.a[i][j] + p.a[i][k] * q.a[k][j] ) % MOD ; } return s ; } node pow(node p,int k) { if( k == 1 ) return p ; node s = pow( p,k/2 ) ; s = mul(s,s) ; if( k%2 ) s = mul(s,p) ; return s ; } int main() { LL n , x , y , i , j , ans ; node p , s ; while( scanf("%I64d %I64d %I64d", &n, &x, &y) != EOF ) { p.n = 4 ; memset(p.a,0,sizeof(p.a)) ; p.a[0][0] = 1 ; p.a[1][0] = p.a[1][1] = (x*x) % MOD ; p.a[2][0] = p.a[2][1] = (2*x*y) % MOD ; p.a[3][0] = p.a[3][1] = (y*y) % MOD ; p.a[1][2] = x % MOD ; p.a[2][2] = y % MOD ; p.a[1][3] = 1 ; s = pow(p,n-1) ; ans = s.a[0][0] * 2 ; for(i = 1 ; i < 4 ; i++) ans = ( ans + s.a[i][0] ) % MOD ; printf("%I64d\n", ans) ; } return 0; }