HDU 5950 Recursive sequence 递推转矩阵

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2882    Accepted Submission(s): 1284

Problem Description

Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers a and b on a blackboard. And then, the cows would say their identity number one by one. The first cow says the first number a and the second says the second number b. After that, the i-th cow says the sum of twice the (i-2)-th number, the (i-1)-th number, and i4. Now, you need to write a program to calculate the number of the N-th cow in order to check if John’s cows can make it right. 

 

Input

The first line of input contains an integer t, the number of test cases. t test cases follow.
Each case contains only one line with three numbers N, a and b where N,a,b < 231 as described above.

 

Output

For each test case, output the number of the N-th cow. This number might be very large, so you need to output it modulo 2147493647.

 

Sample Input

2

3 1 2

4 1 10

 

Sample Output

85

369

 

Hint

In the first case, the third number is 85 = 2*1十2十3^4.
In the second case, the third number is 93 = 2*1十1*10十3^4   and the fourth number is 369 = 2 * 10 十 93 十 4^4.

 

题意     f[n]=f[n-1]+2*f[n-2]+n^4; f[1]=a f[2]=b   求第n项

解析     直接递推肯定会超时的   所以 构造一个7*7系数矩阵 直接快速幂解出来  由于现在比较菜 只会最简单的矩阵 勉强可以写出来。。。。

 

1 2 1 0 0 0 0     　　  f[i-1]　　　　　　　　f[i]

1 0 0 0 0 0 0      　　 f[i-2]　　　　　　　　f[i-1]　　

0 0 1 4 6 4 1　　　　i^4　　　　　　　　  (i+1)^4

0 0 0 1 3 3 1       *   　i^3                  =            (i+1)^3

0 0 0 0 1 2 1　　　　i^2　　　　　　　　  (i+1)^2

0 0 0 0 0 1 1　　　　i　　　　　　　　　   i+1

0 0 0 0 0 0 1　　　   1　　　　　　　　　  1　　

 

AC代码

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const ll mod=,maxn=;
 struct Matrix
 {
     ll m[maxn][maxn];
     Matrix()
     {
         memset(m,,sizeof(m));
     }
     void init()
     {
         for(int i=; i<maxn; i++)
             for(int j=; j<maxn; j++)
                 m[i][j]=(i==j);
     }
     Matrix  operator +(const Matrix &b)const
     {
         Matrix c;
         for(int i=; i<maxn; i++)
         {
             for(int j=; j<maxn; j++)
             {
                 c.m[i][j]=(m[i][j]+b.m[i][j])%mod;
             }
         }
         return c;
     }
     Matrix  operator *(const Matrix &b)const
     {
         Matrix c;
         for(int i=; i<maxn; i++)
         {
             for(int j=; j<maxn; j++)
             {
                 for(int k=; k<maxn; k++)
                 {
                     c.m[i][j]=(c.m[i][j]+(m[i][k]*b.m[k][j])%mod)%mod;
                 }
             }
         }
         return c;
     }
     Matrix  operator^(const ll &t)const
     {
         Matrix ans,a=(*this);
         ans.init();
         ll n=t;
         while(n)
         {
             if(n&) ans=ans*a;
             a=a*a;
             n>>=;
         }
         return ans;
     }
 };
 int main()
 {
     int t;
     ll n,m,a,b;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%lld %lld %lld",&n,&a,&b);
         if(n==)
         {
             cout<<a%mod<<endl;
             continue;
         }
         if(n==)
         {
             cout<<b%mod<<endl;
             continue;
         }
         Matrix temp;
         temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=;temp.m[][]=;
         temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=;temp.m[][]=;
         temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=;temp.m[][]=;
         temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=;temp.m[][]=;
         temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=;temp.m[][]=;
         temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=;temp.m[][]=;
         temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=,temp.m[][]=;temp.m[][]=;
         Matrix aa,bb;
         bb.m[][]=b%mod;
         bb.m[][]=a%mod;
         bb.m[][]=;
         bb.m[][]=;
         bb.m[][]=;
         bb.m[][]=;
         bb.m[][]=;
         aa=temp^(n-);
         aa=aa*bb;
         cout<<aa.m[][]%mod<<endl;
     }
     return ;
 }