HDUOJ--4565 So Easy!

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1729    Accepted Submission(s): 556

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

 

 

此题的思路在于这儿....看图...

 

/*快速矩@ coder Gxjun*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
__int64 ma[2][2];   //matrix
__int64 ans[2][2]; 
void init(__int64 a[][2])
{
    int i;
   for(i=0;i<2;i++)
          a[i][i]=1;   //E 设置为1
          a[1][0]=0;
          a[0][1]=0;
}
int main()
{
    int    a,b,m,n,i,j,k;
    __int64 temp[2] ;
    __int64 hop[2][2];
    while(scanf("%d%d%d%d",&a,&b,&n,&m)!=EOF)
    {
        if(n==1)
        {
            printf("%I64d
",(2*a)%m);
            continue;
        }
        else if(n==2)
        {
            printf("%I64d
",(2*(a*a+b))%m);
            continue;
        }
        n-=2;
        init(ans);
        /*init(ma);*/
         ma[0][0]=(2*a)%m;
         ma[0][1]=(b-a*a)%m;
         ma[1][0]=1;
         ma[1][1]=0;
        while(n>0)
        {
            if(n&1)
            {
                for(k=0;k<2;k++)
                {
                 memset(temp,0,sizeof(temp));
                 for(i=0;i<2;i++)
                 {
                    for(j=0;j<2;j++)
                    {
                      temp[i]+=ans[k][j]*ma[j][i];
                      temp[i]%=m;
                    }
                 }
                 ans[k][0]=temp[0];
                 ans[k][1]=temp[1];
                }
                 n--;
                continue;
            }
           memset(hop,0,sizeof(hop));
            for(k=0;k<2;k++)
            {
              for(i=0;i<2;i++)
              {
                for(j=0;j<2;j++)
                { 
                  hop[k][i]+=ma[k][j]*ma[j][i];
                  hop[k][i]%=m;    
                }
             }
            }
            for(i=0;i<2;i++)
            {
                for(j=0;j<2;j++)
                {
                  ma[i][j]=hop[i][j];
                }
            }
            n>>=1;
        }
        __int64 gong=(2*((ans[0][1]*a)%m+(ans[0][0]*(a*a+b)%m)%m)%m)%m;
        if(gong<0) gong=m+gong;
        printf("%I64d
",gong);
    }
    return 0;
}