HDU 5950 Recursive sequence 【递推+矩阵快速幂】 (2016ACM/ICPC亚洲区沈阳站)

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 249    Accepted Submission(s): 140

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）

 

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题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=5950

题目大意：

　　F_i=F_i-1+2F_i-2+i⁴。给定F₁和F₂求F_n。

题目思路：

　　【递推+矩阵快速幂】

　　现场用算了1个多小时的公式过了。

　　主要还是我太菜。递推写的太少。

　　先考虑f(i)=f(i-1)+2f(i-2)，很容易写出递推矩阵

　　　　0 2

　　　　1 1

　　(i+1)⁴=i⁴+4i³+6i²+4i+1。

　　所以需要在递推矩阵种存下i的4 3 2 1 0次幂，以便推出(i+1)⁴，矩阵为

　　　　1 0 0 0 0

　　　　4 1 0 0 0

　　　　6 3 1 0 0

　　　　4 3 2 1 0

　　　　1 1 1 1 1

　　于是f={f_i-1,f_i,i⁴,i³,i²,i¹,i⁰}，将以上两个矩阵合并，即可推出{f_i,f_i+1,(i+1)⁴,(i+1)³,(i+1)²,(i+1)¹,(i+1)⁰}.矩阵如下

　　　　0 2 0 0 0 0 0

　　　　1 1 0 0 0 0 0

　　　　0 1 1 0 0 0 0

　　　　0 4 4 1 0 0 0

　　　　0 6 6 3 1 0 0

　　　　0 4 4 3 2 1 0

　　　　0 1 1 1 1 1 1

　　推出转移矩阵后只需要根据n求矩阵快速幂即可。

 //
 //by coolxxx
 //#include<bits/stdc++.h>
 #include<iostream>
 #include<algorithm>
 #include<string>
 #include<iomanip>
 #include<map>
 #include<stack>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<memory.h>
 #include<time.h>
 #include<stdio.h>
 #include<stdlib.h>
 #include<string.h>
 //#include<stdbool.h>
 #include<math.h>
 #pragma comment(linker,"/STACK:1024000000,1024000000")
 #define min(a,b) ((a)<(b)?(a):(b))
 #define max(a,b) ((a)>(b)?(a):(b))
 #define abs(a) ((a)>0?(a):(-(a)))
 #define lowbit(a) (a&(-a))
 #define sqr(a) ((a)*(a))
 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
 #define mem(a,b) memset(a,b,sizeof(a))
 #define eps (1e-8)
 #define J 10000
 #define mod 2147493647
 #define MAX 0x7f7f7f7f
 #define PI 3.14159265358979323
 #define N 14
 #define M 7
 using namespace std;
 typedef long long LL;
 double anss;
 LL aans;
 int cas,cass;
 int n,m,lll,ans;
 LL f[N];
 LL a[N][N];
 LL ma[N][N]={{},
     {,,,,,,,},
     {,,,,,,,},
     {,,,,,,,},
     {,,,,,,,},
     {,,,,,,,},
     {,,,,,,,},
     {,,,,,,,}};
 void multi(LL a[][N],LL b[][N],LL c[][N])
 {
     int i,j,k;
     LL t[N][N];
     mem(t,);
     for(i=;i<=M;i++)
         for(j=;j<=M;j++)
             for(k=;k<=M;k++)
                 t[i][j]=(t[i][j]+a[i][k]*b[k][j]%mod)%mod;
     memcpy(c,t,sizeof(t));
 }
 void mi(LL a[][N],int y)
 {
     LL tmp[N][N];
     mem(tmp,);
     tmp[][]=tmp[][]=tmp[][]=tmp[][]=tmp[][]=tmp[][]=tmp[][]=;
     while(y)
     {
         if(y&)multi(tmp,a,tmp);
         y>>=;multi(a,a,a);
     }
     memcpy(a,tmp,sizeof(tmp));
 }
 void work()
 {
     LL t[N];
     mem(t,);
     int i,j;
     for(i=;i<=M;i++)
         for(j=;j<=M;j++)
             t[i]=(t[i]+f[j]*a[j][i]%mod)%mod;
     memcpy(f,t,sizeof(t));
 }
 int main()
 {
     #ifndef ONLINE_JUDGE
     freopen("1.txt","r",stdin);
 //    freopen("2.txt","w",stdout);
     #endif
     int i,j,k;
     int x,y,z;
 //    init();
     for(scanf("%d",&cass);cass;cass--)
 //    for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
 //    while(~scanf("%s",s))
 //    while(~scanf("%d%d",&n,&m))
     {
         memcpy(a,ma,sizeof(a));
         scanf("%d%lld%lld",&n,&f[],&f[]);
         f[]=,f[]=,f[]=,f[]=,f[]=;
         if(n==)
         {
             printf("%lld\n",f[]);
             continue;
         }
         mi(a,n-);
         work();
         printf("%lld\n",f[]);
     }
     return ;
 }
 /*
 //

 //
 */