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): 2042    Accepted Submission(s): 914

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

23 1 24 1 10

 

Sample Output

85369
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：

已知 F(1) = a , F(2) = b , F(n) = F(n-1) + F(n-2) * 2 + n^4, 求F(n)

思路：

构造矩阵。

类似斐波那契数矩阵构造

因为有n^4 所以想到要用其得到 (n+1)^4 = n^4 + 4*n^3 + 6*n^2 + 4*n + 1

可知得到(n+1)^4 需要 n^4 , n^3, n^2 , n , 1

得到F(n)则需要F(n-1)    F(n-2)和n^4

构造矩阵 A(n^4    n^3    n^2    n    1    F(n-1)    F(n-2) ) 

需另外构造矩阵 B 使得 A*B = ((n+1)^4    (n+1)^3    (n+1)^2    n+1    1    F(n)    F(n-1) ) 

得矩阵B

之后用矩阵快速幂

#include<iostream>#include<string.h>#include<stdio.h>#include<math.h>using namespace std;#define LL long longLL T,N,a,b;const LL mod = 2147493647;struct node{    LL a[10][10],h,l;    node(){        memset(a,0,sizeof a);    }    friend node operator * (node a,node b)    {        node c;        c.h = a.h,c.l = b.l;        for(int i=0;i<a.h;i++){            for(int j=0;j<b.l;j++){                for(int k=0;k<7;k++){                    c.a[i][j] += a.a[i][k] * b.a[k][j];                    c.a[i][j] %= mod;                }            }        }        return c;    }};node A,B;node jqkm(node a,LL m){    node base = a,ans;    for(int i=0;i<7;i++){        ans.a[i][i] = 1;    }    ans.h = ans.l = base.h = base.l = 7;    while(m){        if(m&1) ans = ans*base;        base=base*base;        m>>=1;    }    return ans;}int main(){    scanf("%lld",&T);    LL aa[10][10]={{1,4,6,4,1,0,0},{0,1,3,3,1,0,0},{0,0,1,2,1,0,0},{0,0,0,1,1,0,0},{0,0,0,0,1,0,0},{1,0,0,0,0,1,2},{0,0,0,0,0,1,0}};    A.h = A.l = 7;    for(int i=0;i<7;i++){        for(int j=0;j<7;j++){            A.a[i][j] = aa[i][j];        }    }    while(T--){        scanf("%lld%lld%lld",&N,&a,&b);        if(N==1){            printf("%lld\n",a%mod);            continue;        }        else if(N==2){            printf("%lld\n",b%mod);            continue;        }        LL bb[10]={81,27,9,3,1,b%mod,a%mod};        for(int i=0;i<7;i++) B.a[i][0] = bb[i];        B.h=7,B.l=1;        node C = jqkm(A, N-2);        B = C*B;        printf("%lld\n",B.a[5][0]);    }    return 0;}