Another kind of Fibonacci
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3509 Accepted Submission(s): 1401
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 Out
6 196
题意：多组数据。每组数据有N，X，Y，A(0)=1 ,A(1)=1 ,A(N)=X*A(N-1)+Y*A(N-2)(N >= 2),求S(N)=A(0)2+A(1)2+…+A(n)2
题解：利用A(N)=X*A(N-1)+Y*A(N-2)，S(N)=S(N-1)+A(N)2,A(N)2=X2*A(N-1)2+2XY*A(N-1)A(N-2)+Y2*A(N-2)2构造矩阵。
代码：
#include<bits/stdc++.h> using namespace std; const int mod=10007; struct node { long long Martix[4][4]; node operator *(const node&n) const { int i,j,k;node sum; for(i=0;i<4;i++) for(j=0;j<4;j++) { sum.Martix[i][j]=0; for(k=0;k<4;k++) sum.Martix[i][j]=(sum.Martix[i][j]+Martix[i][k]*n.Martix[k][j])%mod; } return sum; } }; int main() { long long n,x,y; node B,ans; while(~scanf("%lld%lld%lld",&n,&x,&y)) { fill(B.Martix[0],B.Martix[0]+16,0); fill(ans.Martix[0],ans.Martix[0]+16,0); for(int i=0;i<4;i++) ans.Martix[i][i]=1; B.Martix[0][0]=B.Martix[1][0]=B.Martix[1][2]=1; B.Martix[1][1]=(x*x)%mod; B.Martix[2][1]=(y*y)%mod; B.Martix[3][1]=(2*x*y)%mod; B.Martix[1][3]=x%mod; B.Martix[3][3]=y%mod; while(n) { if(n&1) ans=ans*B; n>>=1; B=B*B; } printf("%lld\n",(ans.Martix[0][0]+ans.Martix[1][0]+ans.Martix[2][0]+ans.Martix[3][0])%mod); } system("pause"); return 0; }