HDU5950 Recursive sequence —— 矩阵快速幂

题目链接：https://vjudge.net/problem/HDU-5950

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2727    Accepted Submission(s): 1226

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.

 

Source

2016ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

Recommend

jiangzijing2015

 

题意：

求 f(n) = f(n−1) + 2*f(n−2) + n^4，其中 f(1)=a,f(2)=b

题解：

典型的矩阵快速幂的运用。关键是i^4怎么维护？我们可以当成求第i+1项，那么i^4就变成了（i+1）^4。那么这时我们可以用二项式定理从i^4、i^3、i^2、i^1、i^0的组合中得到（i+1）^4。也就是说总共需要维护：f[i+1]、f[i]、（i+1）^4、（i+1）^3、（i+1）^2、（i+1）^1、（i+1）^0。矩阵如下：

代码如下:

 #include <bits/stdc++.h>
 #define rep(i,s,t) for(int (i)=(s); (i)<=(t); (i)++)
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const LL mod = ;
 const int maxn = 1e5;

 struct Mat
 {
     LL mat[][];
     void init()
     {
         rep(i,,) rep(j,,)
             mat[i][j] = (i==j);
     }
 };

 Mat p = {   , , , , , , ,
             , , , , , , ,
             , , , , , , ,
             , , , , , , ,
             , , , , , , ,
             , , , , , , ,
             , , ,  ,, ,
         };

 Mat mul(Mat x, Mat y)
 {
     Mat s;
     ms(s.mat,);
     rep(i,,) rep(j,,) rep(k,,)
         s.mat[i][j] += (x.mat[i][k]*y.mat[k][j])%mod, s.mat[i][j] %= mod;
     return s;
 }

 Mat qpow(Mat x, LL y)
 {
     Mat s;
     s.init();
     while(y)
     {
         if(y&)  s = mul(s, x);
         x = mul(x, x);
         y >>= ;
     }
     return s;
 }
 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--)
     {
         LL n, a, b;
         scanf("%lld%lld%lld",&n,&a,&b);
         if(n == )
         {
             printf("%lld\n",a);
             continue;
         }
         if(n == )
         {
             printf("%lld\n",b);
             continue;
         }

         Mat x = p;
         x = qpow(x, n-);

         LL ans = ;
         ans = (ans + b*x.mat[][]) % mod;
         ans = (ans + a*x.mat[][]%mod) % mod;
         ans = (ans + *x.mat[][]%mod) % mod;
         ans = (ans + *x.mat[][]%mod) % mod;
         ans = (ans + *x.mat[][]%mod) % mod;
         ans = (ans + *x.mat[][]%mod) % mod;
         ans = (ans+x.mat[][]) % mod;
         printf("%lld\n",ans);
     }
 }