HDU 4565 So Easy!

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 < 2 ¹⁵, (a-1) ²< b < a ², 0 < b, n < 2 ³¹.The input will finish with the end of file.

Output

　　For each the case, output an integer S _n.

Sample Input

2 3 1 20132 3 2 2013 2 2 1 2013

Sample Output

414

4

关键是找到递推式

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long use;
use n,m,a,b,x[2][3],y[2][3],c[3][3],e[3][3],d[3][3];

int main()
{
    while (cin>>a>>b>>n>>m)
    {
        x[1][1]=(a+a)%m;    x[1][2]=(2*a*a+2*b)%m;
        memset(y,0,sizeof(y));
        c[1][1]=0;    c[1][2]=(m+(b-a*a)%m)%m;
        c[2][1]=1;    c[2][2]=(a+a)%m;
        e[1][1]=e[2][2]=1;
        e[1][2]=e[2][1]=0;
        for (int u=n-1;u;u>>=1)
        {
            if (u&1) 
            {
                memset(d,0,sizeof(d));
                for (int i=1;i<3;i++)
                    for (int j=1;j<3;j++)
                        for (int k=1;k<3;k++)
                            (d[i][k]+=c[i][j]*e[j][k])%=m;
                memcpy(e,d,sizeof(d));
            };
            memset(d,0,sizeof(d));
            for (int i=1;i<3;i++)
                    for (int j=1;j<3;j++)
                        for (int k=1;k<3;k++)
                            (d[i][k]+=c[i][j]*c[j][k])%=m;
            memcpy(c,d,sizeof(d));
        }
        for (int i=1;i<2;i++)
            for (int j=1;j<3;j++)
                for (int k=1;k<3;k++)
                    (y[i][k]+=x[i][j]*e[j][k])%=m;
        cout<<y[1][1]<<endl;
    }
    return 0;
}