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)。

接下来就可以运用矩阵快速幂求解。

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 long long a,b,n,m;
 5 
 6 struct Mat1{
 7     long long mat[2][2];
 8 };
 9 
10 struct Mat2{
11     long long mat[2][1];
12 };
13 
14 Mat1 operator * (Mat1 A,Mat1 B)
15 {
16     Mat1 ret;
17     memset(ret.mat,0,sizeof(ret));
18     for(int i = 0; i<2; ++i)
19         for(int j = 0; j<2; ++j)
20             for(int k = 0; k<2; ++k)
21                 ret.mat[i][j] = (ret.mat[i][j]+A.mat[i][k]*B.mat[k][j])%m;
22     return ret;
23 }
24 
25 Mat2 operator * (Mat1 A,Mat2 B)
26 {
27     Mat2 ret;
28     memset(ret.mat,0,sizeof(ret));
29     for(int i = 0; i<2; ++i)
30         for(int j = 0; j<1; ++j)
31             for(int k = 0; k<2; ++k)
32                 ret.mat[i][j] = (ret.mat[i][j]+A.mat[i][k]*B.mat[k][j])%m;
33     return ret;
34 }
35 
36 void matquickpowmod()
37 {
38     Mat1 x;
39     x.mat[0][0] = 2*a;
40     x.mat[0][1] = b-a*a;
41     x.mat[1][0] = 1;
42     x.mat[1][1] = 0;
43     Mat2 y;
44     y.mat[0][0] = 2*a;
45     y.mat[1][0] = 2;
46     while(n){
47         if(n&1)
48             y = x*y;
49         x = x*x;
50         n >>= 1;
51     }
52     printf("%I64d
",(y.mat[1][0]%m+m)%m);
53 }
54 
55 int main()
56 {
57     while(~scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&m)){
58         matquickpowmod();
59     }
60     return 0;
61 }