HDU 3802 Ipad,IPhone

Ipad,IPhone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 344    Accepted Submission(s): 126

Problem Description

In ACM_DIY, there is one master called “Lost”. As we know he is a “-2Dai”, which means he has a lot of money.
  
Well, Lost use Ipad and IPhone to reward the ones who solve the following problem.
  
In this problem, we define F( n ) as :
  
Then Lost denote a function G(a,b,n,p) as 

Here a, b, n, p are all positive integer!
If you could tell Lost the value of G(a,b,n,p) , then you will get one Ipad and one IPhone!

 

Input

The first line is one integer T indicates the number of the test cases. (T <= 100)
Then for every case, only one line containing 4 positive integers a, b, n and p.
(1 ≤a, b, n, p≤2*10 ⁹ , p is an odd prime number and a,b < p.)

 

Output

Output one line,the value of the G(a,b,n,p) .

 

Sample Input

   
   
   
   

    
    
    
    4
2 3 1 10007
2 3 2 10007
2 3 3 10007
2 3 4 10007
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    40
392
3880
9941
   
   
   
   

 

题意很简单：就是计算G（a，b，n，p）；

分析：

数据很大

1.数学表达式中有斐波那契数列，但斐波那契数列定义域很大，所以用到了矩阵快速幂求解。

2.斐波那契数列解决后， 还有G表达式，G表达式可分成两部分，前半部分直接快速幂计算。

3.关键是后半部分，可以把他看成一个整体，求出数列特征方程，构建矩阵求解！

4.有一个比较难理解的地方时，取模问题：在求斐波那契数列时是对MOD-1取模，而在求其他数据时是对MOD取余，之所以这么做是因为求斐波那契数列时是对形如A^N取模，根据费马小定理，就是对MOD-1取模，而其他数据则不是A^N这种形式！（这里想了好久！）

细节部分：

关于取模问题，负数取模可以先让其加mod加到正数后再取模，也可以中间直接取模，最后在处理结果（取模）！

代码如下：

#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
ll mod;
struct Mar{
    ll mat[2][2];
};
Mar E={1,0,0,1},unit2={0,1,1,1};
ll my_pow(ll a,ll b){
    ll ans = 1;
    while(b){
        if (b%2)
            ans = (ans*a)%mod;
        b/=2;
        a=(a*a)%mod;
    }
    return ans;
}
Mar mul(Mar a,Mar b){
    ll fumod(ll);
    Mar ans={0,0,0,0};
    for (int i = 0; i < 2; ++i)
        for (int j = 0; j < 2; ++j)
            for (int k = 0; k <2; ++k){
                ans.mat[i][j]+=(a.mat[i][k]*b.mat[k][j])%mod;
                ans.mat[i][j]%=mod;
        }
    return ans;
}
Mar pow2(Mar a,ll n){
    Mar ans = E;
    while(n){
        if (n%2)
            ans = mul(ans,a);
        n/=2;
        a=mul(a,a);
    }
    return ans;
}
ll fib(ll n){
    Mar uu={1,1,1,1};
    Mar p=pow2(unit2,n);
    p=mul(uu,p);
    return p.mat[0][0];
}
int main()
{
    ll T,a,b,n,p;
    cin>>T;
    while(T--){
        cin >> a >> b >> n >> p;
        mod = p;
        ll temp1=(my_pow(a,(p-1)/2)+1)%mod;
        ll temp2=(my_pow(b,(p-1)/2)+1)%mod;
        if (!temp1 || !temp2){cout << "0" << endl;continue;}
        --mod;
        ll fn=fib(n);
        ++mod;
        Mar unit={2,2*(a+b)%mod,1,1};
        Mar tmp1={0,-(a-b)*(a-b)%mod,1,2*(a+b)%mod};
        Mar p=pow2(tmp1,fn);
        p=mul(unit,p);
        ll pn=p.mat[0][0];
        ll ans = pn%mod*temp1%mod*temp2%mod;
        while(ans < 0)ans+=mod;
        cout << ans << endl;
    }
    return 0;
}