HDU/HDOJ 2855 2009 Multi-University Training Contest 5 - Host by NUDT 矩阵二分幂

Fibonacci Check-up

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 615Accepted Submission(s): 333

Problem 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

Source

2009 Multi-University Training Contest 5 - Host by NUDT

有一个公式：

C(N,1)F1+C(N,2)F2+C(N,3)F3+...+C(N,N)FN=F(2N)

证明方法我有点忘了，大致是利用那个比内公式去推

n=0和1最好特判一下

我的代码：

#include<stdio.h> #include<algorithm> #include<string.h> using namespace std; struct mart{ int mat[3][3]; }; int mod; mart kk; mart multi(mart a,mart b){ mart c; int i,j,k; for(i=1;i<=2;i++) for(j=1;j<=2;j++) { c.mat[i][j]=0; for(k=1;k<=2;k++) c.mat[i][j]=(c.mat[i][j]%mod+(a.mat[i][k]%mod)*(b.mat[k][j]%mod)%mod)%mod; } return c; } mart power(int k){ mart p,q; int i,j; for(i=1;i<=2;i++) for(j=1;j<=2;j++) { p.mat[i][j]=kk.mat[i][j]; if(i==j) q.mat[i][j]=1; else q.mat[i][j]=0; } if(k==0) return q; while(k!=1) { if(k&1) { k--; q=multi(p,q); } else { k=k>>1; p=multi(p,p); } } p=multi(p,q); return p; } int main(){ int n,m,tt; scanf("%d",&tt); mart xx; while(tt--) { scanf("%d%d",&n,&m); if(n==0) { printf("0\n"); continue; } if(n==1) { printf("%d\n",1%m); continue; } mod=m; kk.mat[1][1]=0; kk.mat[1][2]=1; kk.mat[2][1]=1; kk.mat[2][2]=1; xx=power(2*n-1); printf("%d\n",(xx.mat[1][1]+xx.mat[1][2])%mod); } return 0; }