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 < 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

 

 

此题的思路在于这儿....看图...

 

 /*快速矩@ coder Gxjun*/
 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 __int64 ma[][];   //matrix
 __int64 ans[][];
 void init(__int64 a[][])
 {
     int i;
    for(i=;i<;i++)
           a[i][i]=;   //E 设置为1
           a[][]=;
           a[][]=;
 }
 int main()
 {
     int    a,b,m,n,i,j,k;
     __int64 temp[] ;
     __int64 hop[][];
     while(scanf("%d%d%d%d",&a,&b,&n,&m)!=EOF)
     {
         if(n==)
         {
             printf("%I64d\n",(*a)%m);
             continue;
         }
         else if(n==)
         {
             printf("%I64d\n",(*(a*a+b))%m);
             continue;
         }
         n-=;
         init(ans);
         /*init(ma);*/
          ma[][]=(*a)%m;
          ma[][]=(b-a*a)%m;
          ma[][]=;
          ma[][]=;
         while(n>)
         {
             if(n&)
             {
                 for(k=;k<;k++)
                 {
                  memset(temp,,sizeof(temp));
                  for(i=;i<;i++)
                  {
                     for(j=;j<;j++)
                     {
                       temp[i]+=ans[k][j]*ma[j][i];
                       temp[i]%=m;
                     }
                  }
                  ans[k][]=temp[];
                  ans[k][]=temp[];
                 }
                  n--;
                 continue;
             }
            memset(hop,,sizeof(hop));
             for(k=;k<;k++)
             {
               for(i=;i<;i++)
               {
                 for(j=;j<;j++)
                 {
                   hop[k][i]+=ma[k][j]*ma[j][i];
                   hop[k][i]%=m;
                 }
              }
             }
             for(i=;i<;i++)
             {
                 for(j=;j<;j++)
                 {
                   ma[i][j]=hop[i][j];
                 }
             }
             n>>=;
         }
         __int64 gong=(*((ans[][]*a)%m+(ans[][]*(a*a+b)%m)%m)%m)%m;
         if(gong<) gong=m+gong;
         printf("%I64d\n",gong);
     }
     return ;
 }