HDU 4686 Arc of Dream

Arc of Dream

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3947    Accepted Submission(s): 1220

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

 

Author

Zejun Wu (watashi)

Source

2013 Multi-University Training Contest 9

 

Recommend

zhuyuanchen520

 

 

由已知条件，可以得到下面矩阵式子

 

进一步变换得到

n>=1,

 

这样，就可以使用矩阵快速幂运算进行求解。

 

#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <functional>
#include <string>
#include <cstring>
#include <queue>
#include <set>
#include <cmath>
#include <cstdio>
using namespace std;
#define IOS ios_base::sync_with_stdio(false)]
typedef long long LL;
const int INF=0x3f3f3f3f;

const int maxn=6;
const int modnum=1000000007;
typedef struct matrix{
    int v[maxn][maxn];
    void init(){memset(v,0,sizeof(v));}
}M;
M mul_mod(const M &a,const M &b,int L,int m,int n)
{
    M c; c.init();
    for(int i=0;i<L;i++){
        for(int j=0;j<n;j++){
            for(int k=0;k<m;k++)//注意j，k范围
                c.v[i][j]=(c.v[i][j]+int((LL)a.v[i][k]*b.v[k][j]%modnum))%modnum;
        }
    }
    return c;
}
M power(M x,int L,LL p)
{
    M tmp; tmp.init();
    for(int i=0;i<L;i++)
        tmp.v[i][i]=1;
    while(p){
        if(p&1) tmp=mul_mod(x,tmp,L,L,L);
        x=mul_mod(x,x,L,L,L);
        p>>=1;
    }
    return tmp;
}
int main()
{
    LL n;
    int a0,ax,ay,b0,bx,by;
    M a,b;
    while(scanf("%lld",&n)!=EOF){
        scanf("%d%d%d%d%d%d",&a0,&ax,&ay,&b0,&bx,&by);
        if(n==0){printf("0
"); continue;}
        a.init();
        a.v[0][0]=1;
        a.v[0][1]=a.v[1][1]=int((LL)ax*bx%modnum);
        a.v[0][2]=a.v[1][2]=int((LL)ax*by%modnum);
        a.v[0][3]=a.v[1][3]=int((LL)ay*bx%modnum);
        a.v[0][4]=a.v[1][4]=int((LL)ay*by%modnum);
        a.v[2][2]=ax%modnum;
        a.v[2][4]=ay%modnum;
        a.v[3][3]=bx%modnum;
        a.v[3][4]=by%modnum;
        a.v[4][4]=1;
        b.init();
        b.v[0][0]=int((LL)a0*b0%modnum);
        b.v[1][0]=int((LL)a0*b0%modnum);
        b.v[2][0]=a0%modnum;
        b.v[3][0]=b0%modnum;
        b.v[4][0]=1;
        a=power(a,5,n-1);
        a=mul_mod(a,b,5,5,1);
        printf("%d
",a.v[0][0]);
    }
}