hdu4565(矩阵快速幂+经典的数学处理)

注意题目的一个关键条件(a-1)²< b < a^{2 ， 于是可以知道}    0 < a-√b < 1 ,所以 (a-√b)^n < 1 . 然后 (a+ √b)^n+(a-√b)^n 的值为整数且正好是 (a+√b)^n的向上取整.

然后就可以得到递推式 f[n+1]=2*a*f[n]-(a*a-b)*f[n-1] .

然后构造矩阵。 幂乘即可.

　　　　　　　　　　　　　　　　So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1364    Accepted Submission(s): 424

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

 

Recommend

zhoujiaqi2010

 

 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
 #include <algorithm>
 #include <math.h>
 #include <map>
 #include <queue>
 #include <sstream>
 #include <iostream>
 using namespace std;
 #define INF 0x3fffffff

 typedef __int64 LL;

 LL a,b,n,m;
 LL g[][];
 LL MOD;

 void cal(LL s[][],LL t[][])
 {
     LL tmp[][];
     memset(tmp,,sizeof(tmp));
     for(int k=;k<;k++)
         for(int i=;i<;i++)
             for(int j=;j<;j++)
             {
                 tmp[i][j]=(tmp[i][j]+s[i][k]*t[k][j])%MOD;
             }
     for(int i=;i<;i++)
         for(int j=;j<;j++)
             s[i][j]=tmp[i][j];
 }

 int main()
 {
     //freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
     //freopen("C:\\Users\\Administrator\\Desktop\\in.txt","w",stdout);
     while(scanf("%d%d%d%d",&a,&b,&n,&m)!=EOF)
     {
         MOD=m;
         g[][]=*a; g[][]=;
         g[][]=b-a*a; g[][]=;

         n--;
         LL sum[][];
         for(int i=;i<;i++)
             for(int j=;j<;j++)
                 if(i==j) sum[i][j]=;
                 else sum[i][j]=;
         while(n)
         {
             if( (n&)!=)
                 cal(sum,g);
             cal(g,g);
             n>>=;
         }
         LL ans=((*a*sum[][]+*sum[][])%MOD+MOD)%MOD;
         cout<<ans<<endl;
     }
     return ;
 }