hdu 4565 So Easy! 矩阵

最新推荐文章于 2020-01-27 10:07:07 发布

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最新推荐文章于 2020-01-27 10:07:07 发布

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分类专栏：矩阵文章标签： HDUJ c++

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本文提供了一种高效的方法来解决特定序列计算问题，通过利用矩阵乘法和模运算简化了计算过程，使得原本复杂的计算变得简单易懂。

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So Easy!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2350 Accepted Submission(s): 729

Problem Description

　　A sequence S_n is defined as:

Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate S_n.
　　You, a top coder, say: So easy!

Input

　　There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 2¹⁵, (a-1)²< b < a², 0 < b, n < 2³¹.The input will finish with the end of file.

Output

　　For each the case, output an integer S_n.

Sample Input

2 3 1 2013
2 3 2 2013
2 2 1 2013

Sample Output

4
14
4

#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
ll a,b,n,m;

struct matrix {
    ll f[2][2];
};

matrix mul(matrix A,matrix B){
    matrix s;
    memset(s.f,0,sizeof s.f);
    ll i,j,k;
    for(k=0;k<2;k++)
    for(i=0;i<2;i++){
        if(!A.f[i][k])  continue;
        for(j=0;j<2;j++){
            if(!B.f[k][j])   continue;
            s.f[i][j]+=A.f[i][k]*B.f[k][j];
            s.f[i][j]%=m;
        }
    }
    return s;
}

matrix pow_mod(matrix A,ll k){
    matrix s;
    s.f[0][0]=s.f[1][1]=1;
    s.f[0][1]=s.f[1][0]=0;
    while(k){
        if(k&1)  s=mul(s,A);
        A=mul(A,A);
        k>>=1;
    }
    return s;
}

int main(){
    while(cin>>a>>b>>n>>m){
        matrix e;
        ll p=2*a;
        ll q=b-a*a;
        e.f[0][0]=p;e.f[0][1]=1;
        e.f[1][0]=q;e.f[1][1]=0;

        e=pow_mod(e,n-1);
        ll ans=((p*e.f[0][0]+2*e.f[1][0])%m+m)%m;
        cout<<ans<<endl;
    }
    return 0;
}