Number Sequence(HDU 1005 构造矩阵 )

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 147964    Accepted Submission(s): 35964

Problem Description

A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

 

 

Sample Input

1 1 3

1 2 1

0
0 0 0

 

Sample Output

2

5

 因为数据量比较大，可以打表找到循环规律，但这种类型的题发现都可以构造矩阵求解

f[n]　　　　　　=  a  b   *   f[n-1]

f[n-1]                  1  0　　  f[n-2]

 

 

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 using namespace std;
 int a,b,n;
 int m[];
 struct Matrix
 {
     int mat[][];
 }p;
 Matrix mul(Matrix a,Matrix b)
 {
     Matrix c;
     for(int i=;i<;i++)
     {
         for(int j=;j<;j++)
         {
             c.mat[i][j]=;
             for(int k=;k<;k++)
                 c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%;
         }
     }
     return c;
 }
 Matrix mod_pow(Matrix x,int n)
 {
     Matrix res;
     memset(res.mat,,sizeof(res.mat));
     for(int i=;i<;i++)
         res.mat[i][i]=;
     while(n)
     {
         if(n&)
             res=mul(res,x);
         x=mul(x,x);
         n>>=;
     }
     return res;
 }
 int main()
 {
     freopen("in.txt","r",stdin);
     while(scanf("%d%d%d",&a,&b,&n)!=EOF)
     {
         if(a==&&b==&&n==)
             break;
         if(n<)
         {
             printf("1\n");
             continue;
         }
         p.mat[][]=a;
         p.mat[][]=;
         p.mat[][]=b;
         p.mat[][]=;
         Matrix ans= mod_pow(p,n-);
         printf("%d\n",(ans.mat[][]+ans.mat[][])%);
     }
 }