HDU 5950 矩阵快速幂

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30    Accepted Submission(s): 20

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）

题意：f[i]=2*f[i-2]+f[i-1]+i^4

题解：[f[i],f[i-1],i^4,i^3,i^2,i,1]

  1 1 0 0 0 0 0

  2 0 0 0 0 0 0

  1 0 1 0 0 0 0

  4 0 4 1 0 0 0

  6 0 6 3 1 0 0

  4 0 4 3 2 1 0

  1 0 1 1 1 1 1

构造矩阵  矩阵快速幂

 /******************************
 code by drizzle
 blog: www.cnblogs.com/hsd-/
 ^ ^    ^ ^
  O      O
 ******************************/
 #include<bits/stdc++.h>
 #include<map>
 #include<set>
 #include<cmath>
 #include<queue>
 #include<bitset>
 #include<math.h>
 #include<vector>
 #include<string>
 #include<stdio.h>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #pragma comment(linker, "/STACK:102400000,102400000")
 using namespace std;
 #define  A first
 #define B second
 const int mod=;
 const int MOD1=;
 const int MOD2=;
 const double EPS=0.00000001;
 typedef __int64 ll;
 const ll MOD=;
 const int INF=;
 const ll MAX=1ll<<;
 const double eps=1e-;
 const double inf=~0u>>;
 const double pi=acos(-1.0);
 typedef double db;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 const ll k=;
 struct matrix
 {
    ll m[][];
 } ans,exm;
 struct matrix matrix_mulit1(struct matrix aa,struct matrix bb)
 {
     struct matrix there;
     for(int i=;i<;i++)
     {
         for(int j=;j<;j++)
         {
             there.m[i][j]=;
             for(int u=;u<;u++)
             there.m[i][j]=(there.m[i][j]+aa.m[i][u]*bb.m[u][j]%k)%k;
         }
     }
     return there;
 }
 struct matrix matrix_mulit2(struct matrix aa,struct matrix bb)
 {
     struct matrix there;
         for(int j=;j<;j++)
         {
             there.m[][j]=;
             for(int u=;u<;u++)
             there.m[][j]=(there.m[][j]+aa.m[][u]*bb.m[u][j]%k)%k;
         }
     return there;
 }
 ll matrix_quick(ll aa,ll bb,ll gg)
 {
 exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;
 exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;
 exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;
 exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;
 exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;
 exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;
 exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;exm.m[][]=;
 ans.m[][]=bb;ans.m[][]=aa;ans.m[][]=;ans.m[][]=;ans.m[][]=;ans.m[][]=;ans.m[][]=;
      gg-=;
      while(gg)
      {
         if(gg&)
         ans=matrix_mulit2(ans,exm);
         exm=matrix_mulit1(exm,exm);
         gg >>= ;
      }
      return ans.m[][];
 }
 int t;
 ll n,a,b;
 int main()
 {
     scanf("%d",&t);
     for(int i=;i<=t;i++)
     {
         scanf("%I64d %I64d %I64d",&n,&a,&b);
         if(n==)
             printf("%I64d\n",a);
         else if(n==)
             printf("%I64d\n",b);
         else
         printf("%I64d\n",(matrix_quick(a, b, n)+k)%k);
     }
     return ;
 }