HDU4686 Arc of Dream

Arc of Dream

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 4473    Accepted Submission(s): 1392

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

11 2 34 5 621 2 34 5 631 2 34 5 6

 

Sample Output

41341902

 

Author

Zejun Wu (watashi)

 

Source

2013 Multi-University Training Contest 9

 

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不难想到矩阵快速幂，然后构造的矩阵。首先

1.ai*bi=（a_i-1*AX+AY）*（b_i-1*BX+BY）展开就知道需要的是什么了。

2.然后要计算的值sum=ai-1*bi-1+ai*bi。然后ai*bi可以又上面的式子得到。

#include <bits/stdc++.h>using namespace std;const int mod = 1e9+7;long long n;long long a0,ax,ay;long long b0,bx,by;struct matrix{    long long a[5][5];    matrix operator * (const matrix & y)const    {        matrix ans= {0};        for(int i=0; i<5; ++i)            for(int j=0; j<5; ++j)            {                for(int k=0; k<5; ++k)ans.a[i][j]+=(a[i][k]*y.a[k][j])%mod;                ans.a[i][j]%=mod;            }        return ans;    }};long long getAns(long long n){    matrix result = {1,0,0,0,0, 0,1,0,0,0, 0,0,1,0,0, 0,0,0,1,0, 0,0,0,0,1};    matrix base = {1,0,0,0,0, (ax*bx)%mod,(ax*bx)%mod,0,0,0, (bx*ay)%mod,(bx*ay)%mod,bx,0,0, (ax*by)%mod,(ax*by)%mod,0,ax,0, (ay*by)%mod,(ay*by)%mod,by,ay,1};    matrix ans= {(a0*b0)%mod,(a0*b0)%mod,b0,a0,1};    while(n)    {        if(n&1ll)result=result*base;        base=base*base;        n>>=1ll;    }    return (ans*result).a[0][0];}int main(){    while(~scanf("%I64d",&n))    {        scanf("%I64d%I64d%I64d",&a0,&ax,&ay);        scanf("%I64d%I64d%I64d",&b0,&bx,&by);        if(!n)        {            puts("0");        }        else        printf("%I64d\n",getAns(n-1));    }    return 0;}