HDU 3802 Ipad,IPhone

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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*10⁹ , 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

42 3 1 100072 3 2 100072 3 3 100072 3 4 10007

Sample Output

4039238809941

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;}

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