HDUOJ--4565 So Easy!

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1729    Accepted Submission(s): 556

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

 

 

此题的思路在于这儿....看图...

 

 1 /*快速矩@ coder Gxjun*/
 2 #include<stdio.h>
 3 #include<string.h>
 4 #include<stdlib.h>
 5 __int64 ma[2][2];   //matrix
 6 __int64 ans[2][2]; 
 7 void init(__int64 a[][2])
 8 {
 9     int i;
10    for(i=0;i<2;i++)
11           a[i][i]=1;   //E 设置为1
12           a[1][0]=0;
13           a[0][1]=0;
14 }
15 int main()
16 {
17     int    a,b,m,n,i,j,k;
18     __int64 temp[2] ;
19     __int64 hop[2][2];
20     while(scanf("%d%d%d%d",&a,&b,&n,&m)!=EOF)
21     {
22         if(n==1)
23         {
24             printf("%I64d\n",(2*a)%m);
25             continue;
26         }
27         else if(n==2)
28         {
29             printf("%I64d\n",(2*(a*a+b))%m);
30             continue;
31         }
32         n-=2;
33         init(ans);
34         /*init(ma);*/
35          ma[0][0]=(2*a)%m;
36          ma[0][1]=(b-a*a)%m;
37          ma[1][0]=1;
38          ma[1][1]=0;
39         while(n>0)
40         {
41             if(n&1)
42             {
43                 for(k=0;k<2;k++)
44                 {
45                  memset(temp,0,sizeof(temp));
46                  for(i=0;i<2;i++)
47                  {
48                     for(j=0;j<2;j++)
49                     {
50                       temp[i]+=ans[k][j]*ma[j][i];
51                       temp[i]%=m;
52                     }
53                  }
54                  ans[k][0]=temp[0];
55                  ans[k][1]=temp[1];
56                 }
57                  n--;
58                 continue;
59             }
60            memset(hop,0,sizeof(hop));
61             for(k=0;k<2;k++)
62             {
63               for(i=0;i<2;i++)
64               {
65                 for(j=0;j<2;j++)
66                 { 
67                   hop[k][i]+=ma[k][j]*ma[j][i];
68                   hop[k][i]%=m;    
69                 }
70              }
71             }
72             for(i=0;i<2;i++)
73             {
74                 for(j=0;j<2;j++)
75                 {
76                   ma[i][j]=hop[i][j];
77                 }
78             }
79             n>>=1;
80         }
81         __int64 gong=(2*((ans[0][1]*a)%m+(ans[0][0]*(a*a+b)%m)%m)%m)%m;
82         if(gong<0) gong=m+gong;
83         printf("%I64d\n",gong);
84     }
85     return 0;
86 }

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