hdu4686 Arc of Dream

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

Total Submission(s): 3354    Accepted Submission(s): 1055

Problem Description

An Arc of Dream is a curve defined by following function:



where

a₀ = A0

a_i = a_i-1*AX+AY

b₀ = B0

b_i = b_i-1*BX+BY

What is the value of AoD(N) modulo 1,000,000,007?

 

Input

There are multiple test cases. Process to the End of File.

Each test case contains 7 nonnegative integers as follows:

N

A0 AX AY

B0 BX BY

N is no more than 10¹⁸, and all the other integers are no more than 2×10⁹.

 

Output

For each test case, output AoD(N) modulo 1,000,000,007.

 

Sample Input

1
1 2 3
4 5 6
2
1 2 3
4 5 6
3
1 2 3
4 5 6

 

Sample Output

4
134
1902

 

这题可以用矩阵快速幂做，先退出公式f[n]=f[n-1]+a[n-1]*b[n-1],

我们可以构造矩阵【f[n-1],a[n-1]*b[n-1],a[n-1],b[n-1],ay,by,ay*by】*A=【f[n],a[n]*b[n],a[n],b[n],ay,by,ay*by】，容易得到

1   0         0       0       0    0   0 

     1 ax*bx   0       0       0    0   0 

     0 ax*by   ax     0       0    0   0 

A=0 ay*bx   0       bx     0    0   0

     0   0         1       0       1    0   0 

     0   0         0       1       0    1   0 

0   1         0       0       0    0   1

然后直接求就行了。

#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 unsigned long long ll;
#define inf 99999999
#define pi acos(-1.0)
#define MOD 1000000007
struct matrix{
    ll n,m,i;
    ll data[9][9];
    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,m,i,j;
    ll ax,bx,ay,by,a0,b0,a1,b1;
    while(scanf("%llu",&n)!=EOF)
    {
        scanf("%llu%llu%llu",&a0,&ax,&ay);
        scanf("%llu%llu%llu",&b0,&bx,&by);
        if(n==0){
            printf("0\n");continue;

        }
        matrix a;
        a.n=a.m=7;
        memset(a.data,0,sizeof(a.data));
        a.data[0][0]=a.data[1][0]=a.data[4][2]=a.data[4][4]=a.data[5][3]=a.data[5][5]=a.data[6][1]=a.data[6][6]=1;
        a.data[1][1]=ax*bx%MOD;
        a.data[2][1]=ax*by%MOD;
        a.data[2][2]=ax%MOD;
        a.data[3][1]=ay*bx%MOD;
        a.data[3][3]=bx%MOD;
        matrix cnt;
        cnt=fast_mod(a,n-1);

        matrix ant;
        ant.n=1;ant.m=7;
        a1=((a0*ax)%MOD+ay)%MOD;
        b1=((b0*bx)%MOD+by)%MOD;
        ant.data[0][0]=a0*b0%MOD;
        ant.data[0][1]=a1*b1%MOD;
        ant.data[0][2]=a1%MOD;
        ant.data[0][3]=b1%MOD;
        ant.data[0][4]=ay%MOD;
        ant.data[0][5]=by%MOD;
        ant.data[0][6]=ay*by%MOD;

        matrix juzhen;
        juzhen=multi(ant,cnt);

        printf("%llu\n",juzhen.data[0][0]%MOD);
    }
    return 0;
}