HDOJ 题目2855 Fibonacci Check-up（矩阵快速幂）

Fibonacci Check-up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1173    Accepted Submission(s): 661


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

 

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 思路：http://blog.csdn.net/xuzengqiang/article/details/7645020
在这里顺便贴下斐波那契数列的通项公式：



 我们可以利用这个公式来简化很多运算：例HDU 2855，意思是输入n,m，求

%m的结果。

具体推导过程如下：

 


 

 

 

 

 

 

 

 

 

ac代码

#include<stdio.h>
#include<string.h>
#include<math.h>
struct s
{
int m[2][2];
};
int n,mm;
struct s muti(struct s a,struct s b)
{
int i,j,k;
struct s c;
memset(c.m,0,sizeof(c.m));
for(i=0;i<2;i++)
{
for(j=0;j<2;j++)
{
for(k=0;k<2;k++)
{
c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mm;
}
c.m[i][j]%=mm;
}
}
return c;
}
struct s powm(struct s a,int n)
{
struct s b;
memset(b.m,0,sizeof(b.m));
b.m[0][0]=b.m[1][1]=1;
while(n)
{
if(n&1)
b=muti(b,a);
a=muti(a,a);
n/=2;
}
return b;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
//int n,m;
struct s e,ans;
e.m[0][0]=e.m[0][1]=e.m[1][0]=1;
e.m[1][1]=0;
scanf("%d%d",&n,&mm);
if(n==0)
{
printf("0\n");
continue;
}
n*=2;
ans=powm(e,n);
printf("%d\n",ans.m[0][1]);
}
}