hdu So Easy!

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2026    Accepted Submission(s): 624

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 < 2 ¹⁵, (a-1) ²< b < a ², 0 < b, n < 2 ³¹.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

 

 1 #include<iostream>

 2 #include<stdio.h>

 3 #include<cstring>

 4 #include<cstdlib>

 5 using namespace std;

 6 typedef __int64 LL;

 7 

 8 LL  p ;

 9 struct Matrix

10 {

11     LL mat[3][3];

12     void init()

13     {

14         mat[1][1]=1;mat[1][2]=0;

15         mat[2][1]=0;mat[2][2]=1;

16     }

17     void mem(LL a,LL b)

18     {

19         mat[1][1]=(2*a)%p; mat[1][2]=(b-a*a)%p;

20         mat[2][1]=1;     mat[2][2]=0;

21     }

22 };

23 Matrix multiply(Matrix cur,Matrix ans)

24 {

25     Matrix now;

26     memset(now.mat,0,sizeof(now.mat));

27     int i,j,k;

28     for(i=1;i<=2;i++)

29     {

30         for(k=1;k<=2;k++)

31         {

32             for(j=1;j<=2;j++)

33             {

34                 now.mat[i][j]+=cur.mat[i][k]*ans.mat[k][j];

35                 now.mat[i][j]%=p;

36                 while(now.mat[i][j]<0) now.mat[i][j]+=p;

37             }

38         }

39     }

40     return now;

41 }

42 void pow_mod(Matrix cur,LL n,LL a,LL b)

43 {

44     Matrix ans;

45     ans.init();

46     while(n)

47     {

48         if(n&1) ans=multiply(ans,cur);

49         n=n>>1;

50         cur=multiply(cur,cur);

51     }

52     LL sum=(ans.mat[1][1]*2*a+ans.mat[1][2]*2)%p;

53     printf("%I64d\n",sum);

54 }

55 int main()

56 {

57     LL a,b,n;

58     while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&p)>0)

59     {

60         Matrix hxl;

61         hxl.mem(a,b);

62         if(n>1)

63         pow_mod(hxl,n-1,a,b);

64         else printf("%I64d\n",(2*a)%p);

65     }

66     return 0;

67 }