Fibonacci Check-up

这道题是Fibonacci 的变形

首先尝试算出1到10大的答案通过找规律，发现他也是Fibonacci 序列

1 3 8 21 55这些个数也都是斐波那契数啊，而且Sn=F2*n，比如S3=8=F6............然后再敲，1A。

构建矩阵是要这样构建的：

E - Fibonacci Check-up Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit  Status

Description

Every ALPC has his own alpc-number just like alpc12, alpc55, alpc62 etc.  As more and more fresh man join us. How to number them? And how to avoid their alpc-number conflicted?  Of course, we can number them one by one, but that’s too bored! So ALPCs use another method called Fibonacci Check-up in spite of collision.  First you should multiply all digit of your studying number to get a number n (maybe huge).  Then use Fibonacci Check-up!  Fibonacci sequence is well-known to everyone. People define Fibonacci sequence as follows: F(0) = 0, F(1) = 1. F(n) = F(n-1) + F(n-2), n>=2. It’s easy for us to calculate F(n) mod m.  But in this method we make the problem has more challenge. We calculate the formula  , is the combination number. The answer mod m (the total number of alpc team members) is just your alpc-number.

Input

First line is the testcase T.  Following T lines, each line is two integers n, m ( 0<= n <= 10^9, 1 <= m <= 30000 )

Output

Output the alpc-number.

Sample Input

2 1 30000 2 30000

Sample Output

1 3

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 #include <iostream> #include <cstdio> #include <cstring> using namespace std; struct matrix { int ma[3][3]; }; int mod; matrix mult(matrix a,matrix b) { matrix ans; memset(ans.ma,0,sizeof(ans.ma)); for(int i=1;i<=2;i++) { for(int j=1;j<=2;j++) { for(int k=1;k<=2;k++) { ans.ma[i][j]+=(a.ma[i][k]*b.ma[k][j])%mod; } ans.ma[i][j]%=mod; } } return ans; } matrix pow(matrix x,int n) { matrix ans; memset(ans.ma,0,sizeof(ans.ma)); for(int i=1;i<=2;i++) { ans.ma[i][i]=1; } while(n) { if(n&1) { ans=mult(ans,x); } x=mult(x,x); n>>=1; } return ans; } int main() { int t,n; scanf("%d",&t); while(t--) { scanf("%d %d",&n,&mod); if(n==0) { printf("0\n"); continue; } matrix sum; sum.ma[1][1]=sum.ma[1][2]=sum.ma[2][1]=1; sum.ma[2][2]=0; sum=pow(sum,2*n); printf("%d\n",sum.ma[1][2]); } return 0; }