2013长沙邀请赛A So Easy!(矩阵快速幂，共轭)

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2286    Accepted Submission(s): 710

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

挑战程序设计竞赛P268，关键是求初始矩阵：

an+1 = a*an+b*bn;

bn+1 = an+a*bn;

a[0][0],a[0][1]分别是公式1里面的系数，a和b;

a[1][0],a[1][1]分别是公式2里面的系数，1和a;

然后就是矩阵快速幂了：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<vector>
 #include<algorithm>
 #include<cmath>
 #define M(a,b) memset(a,b,sizeof(a))
 typedef long long LL;

 using namespace std;

 long long a,b,n,m;

 struct matrix
 {
     LL mat[][];
     void init()
     {
         mat[][] = a;
         mat[][] = b;
         mat[][] = ;
         mat[][] = a;
     }
 };

 matrix mamul(matrix aa,matrix bb)
 {
     matrix c;
     for(int i = ;i<;i++)
     {
         for(int j = ;j<;j++)
         {
             c.mat[i][j] = ;
             for(int k = ;k<;k++)
                 c.mat[i][j]+=(aa.mat[i][k]*bb.mat[k][j]);
             c.mat[i][j]%=m;
         }
     }
     return c;
 }

 matrix mul(matrix s, int k)
 {
     matrix ans;
     ans.init();
     while(k>=)
     {
         if(k&)
             ans = mamul(ans,s);
         k = k>>;
         s = mamul(s,s);
     }
     return ans;
 }

 int main()
 {
     while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&m)==)
     {
             matrix ans;
             ans.init();
             ans = mul(ans,n-);
             printf("%I64d\n",(ans.mat[][]*+m)%m);
     }
     return ;
 }