So Easy! - HUD 4565 递推→矩阵快速幂

So Easy!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2089 Accepted Submission(s): 646



Problem Description

　　A sequence Sn 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 Sn.

　　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 Sn.



Sample Input

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

Sample Output

4
14
4

思路：所有的n次方我们都可以把它看成是a
+b
*根号b，然而a[n+1]=a*a
+b*b
,b[n+1]=a
+a*b[b]。所以我们可以写出一个矩阵 a b 然后通过矩阵的相乘可以得到n次方式 a
5*b
。

1 a b
a

然后根据数论，因为由a和b的关系可以得到，这个答案应该是2*a。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
struct node
{ long long mat[2][2];
};
node mat[4];
long long a,b,n,m,ans;
int b2[100],len;
void mul(int p)
{ int i,j,k;
  memset(mat[3].mat,0,sizeof(mat[3].mat));
  for(i=0;i<=1;i++)
   for(j=0;j<=1;j++)
    for(k=0;k<=1;k++)
     mat[3].mat[i][j]=(mat[3].mat[i][j]+mat[1].mat[i][k]*mat[p].mat[k][j])%m;
  for(i=0;i<=1;i++)
   for(j=0;j<=1;j++)
    mat[1].mat[i][j]=mat[3].mat[i][j];
}
void solve()
{ len=0;
  memset(b2,0,sizeof(b2));
  while(n)
  { len++;
    if(n%2==1)
     b2[len]=1;
    n/=2;
  }
  while(--len)
  { if(b2[len]==0)
     mul(1);
    else
    { mul(1);
      mul(2);
    }
  }
}
int main()
{ int i,j,k;
  while(~scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&m))
  { mat[1].mat[0][0]=mat[2].mat[0][0]=a;
    mat[1].mat[0][1]=mat[2].mat[0][1]=b;
    mat[1].mat[1][0]=mat[2].mat[1][0]=1;
    mat[1].mat[1][1]=mat[2].mat[1][1]=a;
    solve();
    ans=mat[1].mat[0][0]*2%m;
    printf("%I64d\n",ans);
  }
}