HDU 4565 So Easy!

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3459    Accepted Submission(s): 1113

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

 

 

 

解析：矩阵快速幂。

根据共轭式的原理，(a+√b)ⁿ+(a-√b)ⁿ必为整数。

而(a-1)²< b < a²

=> a-1< √b < a

=> 0< a-√b <1

=> 0< (a-√b)ⁿ <1

所以(a+√b)ⁿ+(a-√b)ⁿ = ceil( (a+√b)ⁿ )。

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" alt="" />

接下来就可以运用矩阵快速幂求解。

 

 

 

 #include <cstdio>
 #include <cstring>

 long long a,b,n,m;

 struct Mat1{
     long long mat[][];
 };

 struct Mat2{
     long long mat[][];
 };

 Mat1 operator * (Mat1 A,Mat1 B)
 {
     Mat1 ret;
     memset(ret.mat,,sizeof(ret));
     for(int i = ; i<; ++i)
         for(int j = ; j<; ++j)
             for(int k = ; k<; ++k)
                 ret.mat[i][j] = (ret.mat[i][j]+A.mat[i][k]*B.mat[k][j])%m;
     return ret;
 }

 Mat2 operator * (Mat1 A,Mat2 B)
 {
     Mat2 ret;
     memset(ret.mat,,sizeof(ret));
     for(int i = ; i<; ++i)
         for(int j = ; j<; ++j)
             for(int k = ; k<; ++k)
                 ret.mat[i][j] = (ret.mat[i][j]+A.mat[i][k]*B.mat[k][j])%m;
     return ret;
 }

 void matquickpowmod()
 {
     Mat1 x;
     x.mat[][] = *a;
     x.mat[][] = b-a*a;
     x.mat[][] = ;
     x.mat[][] = ;
     Mat2 y;
     y.mat[][] = *a;
     y.mat[][] = ;
     while(n){
         if(n&)
             y = x*y;
         x = x*x;
         n >>= ;
     }
     printf("%I64d\n",(y.mat[][]%m+m)%m);
 }

 int main()
 {
     while(~scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&m)){
         matquickpowmod();
     }
     return ;
 }