【HDU3306】Another kind of Fibonacci
Another kind of Fibonacci
题目描述
As we all known , the Fibonacci series : \(F\)(\(0\))\(=1\),\(F\)(\(1\))\(=1\),\(F\)(\(N\))\(=F\)(\(N-1\))\(+F\)(\(N-2\))(\(N \ge 2\)).Now we define another kind of Fibonacci : \(A\)(\(0\))\(=1\),\(A\)(\(1\))\(=1\),\(A\)(\(N\))\(=X *A\)(\(N-1\))\(+Y*A\)(\(N-2\))(\(N \ge 2\)).And we want to Calculate \(S\)(\(N\)),\(S\)(\(N\))\(=A\)(\(0\))\(^2+A\)(\(1\))\(^2+\)……\(+A\)(\(n\))\(^2\).
输入格式
There are several test cases.
Each test case will contain three integers,\(N\),\(X\),\(Y\).
\(N\):\(2 \le N \le 2^{31}-1\)
\(X\):\(2 \le X \le 2^{31}-1\)
\(Y\):\(2 \le Y \le 2^{31}-1\)
输出格式
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.
样例输入
2 1 1 3 2 3
样例输出
6 196
题解
这道题如果只是求\(f\)(\(n\))的话就和板题一样简单了,但是他要求的是\(f(n)\)的平方的前缀和。
那么我们接下来开始构造矩阵:
首先很容易想到\(S\)(\(n\))\(=S\)(\(n-1\))+\(f\)(\(n\))\(^2\)
那么我们的矩阵里现在有\(S\)(\(n\)),\(f\)(\(n\))\(^2\)两个元素。
但是我们无法转移\(f\)(\(n\))\(^2\),
所以我们把\(f\)(\(n\))\(^2\)拆开来,
\(f\)(\(n\))\(^2=x^2*f\)(\(n-1\))\(^2+2*x*y*f\)(\(n-1\))\(*f\)(\(n-2\))\(+y^2*f\)(\(n-2\))\(^2\)
我们又考虑到\(f\)(\(n\))\(*f\)(\(n-1\))\(=\)(\(x*f\)(\(n-1\))\(+y*f\)(\(n-2\)))\(*f\)(\(n-1\))\(=x*f\)(\(n-1\))\(^2+y*f\)(\(n-1\))\(*f\)(\(n-2\))
那么我们再往矩阵里加上\(f\)(\(n-1\))\(^2\),\(f\)(\(n-1\))\(*f\)(\(n-2\)),就能完成转移了。
构造完后矩阵如下:
上面写的\(f\)(\(n\))就是题目中的\(A\)(\(n\)),习惯性地写成了\(f\)(\(n\)),写完后也懒得改了,大家能理解就好了。
上代码:
#include<bits/stdc++.h>
#define mod 10007
using namespace std;
int n,x,y;
struct aa{
int a[9][9];
};
aa u;
aa cc(aa x,aa y){
aa ans;
for(int j=1;j<=4;j++)
for(int i=1;i<=4;i++){
ans.a[j][i]=0;
for(int o=1;o<=4;o++)
ans.a[j][i]=(ans.a[j][i]+x.a[j][o]*y.a[o][i])%mod;
}
return ans;
}
aa ksm(aa x,int p){
aa ans;
for(int j=1;j<=4;j++)
for(int i=1;i<=4;i++)
ans.a[j][i]=(j==i);
while(p){
if(p&1) ans=cc(ans,x);
x=cc(x,x);
p>>=1;
}
return ans;
}
int main(){
u.a[1][1]=u.a[3][2]=1;
while(scanf("%d%d%d",&n,&x,&y)!=EOF){
if(n==0){puts("1");continue;}
if(n==1){puts("2");continue;}
x%=mod;y%=mod;
u.a[1][2]=(x*x)%mod;
u.a[1][3]=(y*y)%mod;
u.a[1][4]=(2*x*y)%mod;
u.a[2][2]=(x*x)%mod;
u.a[2][3]=(y*y)%mod;
u.a[2][4]=(2*x*y)%mod;
u.a[4][2]=x;
u.a[4][4]=y;
aa ans=ksm(u,n-1);
int anss=0;
for(int j=2;j<=4;j++)
anss=(anss+ans.a[1][j])%mod;
printf("%d\n",(anss+ans.a[1][1]*2)%mod);
}
return 0;
}