hdu3306 Another kind of Fibonacci

来源：互联网发布：华为p9软件下载编辑：程序博客网时间：2024/05/19 18:15

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2046 Accepted Submission(s): 813

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

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

Author

wyb

这题可以用矩阵快速幂做，构造【A[n-2],A[n-1],A[n-1]*A[n-1],A[n-1]*A[n-2],A[n-2]*A[n-2],S[n-1]】*A=【A[n-1],A[n],A[n]*A[n],A[n]*A[n-1],A[n-1]*A[n-1],S[n]】.

0 y 0 0 0 0

1 x 0 0 0 0

其中A=0 0 x*x x 1 x*x

0 0 2*x*y y 0 2*x*y

0 0 y*y 0 0 y*y

0 0 0 0 0 1

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;typedef long long ll;#define inf 99999999#define pi acos(-1.0)#define MOD 10007struct matrix{    ll n,m,i;    ll data[8][8];    void init_danwei(){        for(i=0;i<n;i++){            data[i][i]=1;        }    }};matrix multi(matrix &a,matrix &b){    ll i,j,k;    matrix temp;    temp.n=a.n;    temp.m=b.m;    for(i=0;i<temp.n;i++){        for(j=0;j<temp.m;j++){            temp.data[i][j]=0;        }    }    for(i=0;i<a.n;i++){        for(k=0;k<a.m;k++){            if(a.data[i][k]>0){                for(j=0;j<b.m;j++){                    temp.data[i][j]=(temp.data[i][j]+(a.data[i][k]*b.data[k][j])%MOD )%MOD;                }            }        }    }    return temp;}matrix fast_mod(matrix &a,ll n){    matrix ans;    ans.n=a.n;    ans.m=a.m;    memset(ans.data,0,sizeof(ans.data));    ans.init_danwei();    while(n>0){        if(n&1)ans=multi(ans,a);        a=multi(a,a);        n>>=1;    }    return ans;}int main(){    ll n,x,y,m,i,j;    while(scanf("%lld%lld%lld",&n,&x,&y)!=EOF)    {        x=x%MOD;        y=y%MOD;        matrix a;        a.n=a.m=6;        memset(a.data,0,sizeof(a.data));        a.data[0][1]=a.data[3][3]=y;        a.data[1][1]=a.data[2][3]=x;        a.data[2][4]=a.data[1][0]=a.data[5][5]=1;        a.data[2][2]=a.data[2][5]=x*x%MOD;        a.data[4][2]=a.data[4][5]=y*y%MOD;        a.data[3][2]=a.data[3][5]=2*x*y%MOD;        matrix ans;        ans=fast_mod(a,n-1);        matrix cnt;        cnt.n=1;cnt.m=6;        cnt.data[0][0]=cnt.data[0][3]=cnt.data[0][4]=1;        cnt.data[0][1]=cnt.data[0][2]=1;        cnt.data[0][5]=2;        matrix ant;        ant=multi(cnt,ans);        printf("%lld\n",ant.data[0][5]);    }    return 0;}

0 0