Hduoj2855【矩阵数学】
/*Fibonacci Check-up
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1257 Accepted Submission(s): 708
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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*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int n, m, ans[3][3], temp[3][3];
void mul(int a[3][3], int b[3][3], int flag)
{
int temp1[3][3] = {0};//保存中间结果
for(int i = 1; i < 3; ++i)
for(int j = 1; j < 3; ++j)
{
for(int k = 1; k < 3; ++k)
temp1[i][j] = (temp1[i][j] + a[i][k] * b[k][j]) % m;
}
if(flag == 1)//更新
{
ans[1][1] = temp1[1][1];
ans[1][2] = temp1[1][2];
ans[2][1] = temp1[2][1];
ans[2][2] = temp1[2][2];
}
else
{
temp[1][1] = temp1[1][1];
temp[1][2] = temp1[1][2];
temp[2][1] = temp1[2][1];
temp[2][2] = temp1[2][2];
}
}
void matrix_pow()
{
ans[1][1] = 1; ans[1][2] = 0;
ans[2][1] = 0; ans[2][2] = 1;
while(n)
{
if(n & 1)
mul(ans, temp, 1);
mul(temp, temp, 2);
n >>= 1;
}
}
int main()
{
int i, j, k, t;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &n, &m);
n *= 2;
if(n == 0)
{
printf("0\n");
continue;
}
temp[1][1] = 1;temp[1][2] = 1;
temp[2][1] = 1;temp[2][2] = 0;
matrix_pow();
printf("%d\n", ans[1][2] % m);
}
return 0;
}
题意：给出n和m求公式
的值。
思路：这里有2个难点，第一个就是公式化简，第二个就是斐波那契的矩阵乘法的规律和矩阵的快速幂。先说公式化简吧~下面附上 图片
公式化简以后就要用到斐波那契的矩阵乘法规律：
我们可以先保存b=f(1),a=f(0)，然后每次设： b'=a+b a'=b。后利用a'和b'一直循环即可。同时我们可以将b a看做一个向量[b a]，前面的操作就可以乘以矩阵：
|1 1|*[b a]=[a+b b]。
|1 0|
也就是说，如果我们要求第100个fibonacci数，只需要将矩阵[1 0]乘上
1 1
1 0
的一百次方，再取出第二项即可。
1 0
难点：首先在于公式的化简，其次就是斐波那契的矩阵求法从而用矩阵快速幂求得结果。