hdu So Easy!

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2026    Accepted Submission(s): 624

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

 

Source

2013 ACM-ICPC长沙赛区全国邀请赛——题目重现

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef __int64 LL;

LL  p ;
struct Matrix
{
    LL mat[3][3];
    void init()
    {
        mat[1][1]=1;mat[1][2]=0;
        mat[2][1]=0;mat[2][2]=1;
    }
    void mem(LL a,LL b)
    {
        mat[1][1]=(2*a)%p; mat[1][2]=(b-a*a)%p;
        mat[2][1]=1;     mat[2][2]=0;
    }
};
Matrix multiply(Matrix cur,Matrix ans)
{
    Matrix now;
    memset(now.mat,0,sizeof(now.mat));
    int i,j,k;
    for(i=1;i<=2;i++)
    {
        for(k=1;k<=2;k++)
        {
            for(j=1;j<=2;j++)
            {
                now.mat[i][j]+=cur.mat[i][k]*ans.mat[k][j];
                now.mat[i][j]%=p;
                while(now.mat[i][j]<0) now.mat[i][j]+=p;
            }
        }
    }
    return now;
}
void pow_mod(Matrix cur,LL n,LL a,LL b)
{
    Matrix ans;
    ans.init();
    while(n)
    {
        if(n&1) ans=multiply(ans,cur);
        n=n>>1;
        cur=multiply(cur,cur);
    }
    LL sum=(ans.mat[1][1]*2*a+ans.mat[1][2]*2)%p;
    printf("%I64d
",sum);
}
int main()
{
    LL a,b,n;
    while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&p)>0)
    {
        Matrix hxl;
        hxl.mem(a,b);
        if(n>1)
        pow_mod(hxl,n-1,a,b);
        else printf("%I64d
",(2*a)%p);
    }
    return 0;
}