hdu3306之矩阵快速幂
Another kind of Fibonacci
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1208 Accepted Submission(s): 462
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 <= 2 31 – 1
X : 2<= X <= 2 31– 1
Y : 2<= Y <= 2 31 – 1
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
Author
wyb
由于递推公式求值可以用矩阵快速幂求值，所以在这很容易想到运用矩阵快速幂求s[n],现在问题是如何去求A[n]^2
由题目A(N) = X * A(N - 1) + Y * A(N - 2) => A[n]^2=x^2*A[n-1]^2+y^2*A[n-2]^2+2*x*y*A[n-1]*A[n-2],
A[n]也是递推公式，也可以用矩阵快速幂求值，现在问题又转化为如何求A[n-1]*A[n-2]
由题目A(N) = X * A(N - 1) + Y * A(N - 2) =〉A[n]*A[n-1]=x*A[n-1]^2+y*A[n-1]*A[n-2],
A[n-1]*A[n-2]也是递推公式，所以也可以用矩阵快速幂求值
所以综合可以得到初始矩阵:
1 1 0 0 s[n] s[n+1]
0 x^2 y^2 2xy A[n+1]^2 A[n+2]^2
0 1 0 0 * A[n]^2 = A[n+1]^2
0 x 0 y A[n+1]*A[n] A[n+2]*A[n+1]
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;
const int MAX=4;
const int mod=10007;
int array[MAX][MAX],sum[MAX][MAX];
void MatrixMult(int a[MAX][MAX],int b[MAX][MAX]){
int c[MAX][MAX]={0};
for(int i=0;i<MAX;++i){
for(int j=0;j<MAX;++j){
for(int k=0;k<MAX;++k){
c[i][j]+=a[i][k]*b[k][j];
}
}
}
for(int i=0;i<MAX;++i){
for(int j=0;j<MAX;++j)a[i][j]=c[i][j]%mod;
}
}
int MatrixPow(int k){
for(int i=0;i<MAX;++i){
for(int j=0;j<MAX;++j)sum[i][j]=(i == j);
}
while(k){
if(k&1)MatrixMult(sum,array);
MatrixMult(array,array);
k>>=1;
}
return (sum[0][0]+sum[0][1]+sum[0][2]+sum[0][3])%mod;//s[0]=,A[0]^2=1
}
int main(){
int n,x,y;
while(scanf("%d%d%d",&n,&x,&y)!=EOF){
x=x%mod,y=y%mod;
array[0][0]=array[0][1]=1,array[0][2]=array[0][3]=0;
array[1][0]=0,array[1][1]=(x*x)%mod,array[1][2]=(y*y)%mod,array[1][3]=(2*x*y)%mod;
array[2][0]=array[2][2]=array[2][3]=0,array[2][1]=1;
array[3][0]=array[3][2]=0,array[3][1]=x,array[3][3]=y;
cout<<MatrixPow(n)<<endl;
}
return 0;
}