poj 3070 題解 矩陣乘法

【序言】驚奇的發現，矩陣乘法真是個優化程序的好東西。像矩陣乘法啊、堆啊，我會陸續學習。

【介紹】矩陣乘法：設A矩陣大小m*p，b矩陣大小為p*n，且C=A*B，那麼C矩陣大小為m*n。C數組中的c[i][j]表示A矩陣的第i行和b矩陣的第j列兩兩相乘的和。矩陣具有結合律，但不具有交換律。

【例題*poj3070】

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n ? 1} + F_{n ? 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

.<喎?http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+R2l2ZW4gYW4gaW50ZWdlciA8ZW0+bjwvZW0+LCB5b3VyIGdvYWwgaXMgdG8gY29tcHV0ZSB0aGUgbGFzdCA0IGRpZ2l0cyBvZiA8ZW0+RjxzdWI+bjwvc3ViPjwvZW0+LjwvcD4KCjxwIGNsYXNzPQ=="pst">Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number ?1.

Output

For each test case, print the last four digits of F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

.

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

.

Source

【分析】矩陣乘法重在推遞推式。由於初學，我只會做這種簡單的題目。剛開始自己是這麼推的。設A矩陣是1*2，元素是（a,b），B矩陣是2*2，元素是（0,1,1,1），這樣我們可以計算出，A*B=（b,a+b)。這就起到了遞推的效果。可以想象，再乘一個B，結果就是（a+b,a+2b)。這就是一個迭代。

然而編完後，我發現我的程序有點問題。我是先算B^n（用快速冪），再算與A的積。但是因為矩陣不具有交換律，所以得到的結果可能有問題。這是我才驚奇的發現，題目上有遞推式！然後套用一下，A了。

【原代碼（有bug）】

#include
using namespace std;
const int mo=10000;
struct arr{int v[3][3];}a,b;
int n;
arr chen(arr x,arr y,int m,int n,int p)
{
  arr s;
  for (int i=1;i<=m;i++)
    for (int j=1;j<=p;j++)
      {
        s.v[i][j]=0;
        for (int k=1;k<=n;k++)
          s.v[i][j]=(s.v[i][j]+x.v[i][k]*y.v[k][j]%mo)%mo;
      }
  return s;
}
void quick()
{
  int mi=n-1;arr res=b;
  while (mi>0)
  {
    if (mi&1) res=chen(res,b,2,2,2);
    b=chen(b,b,2,2,2);
    mi/=2;
  }
  a=chen(a,res,1,2,2);
}
int main()
{
  while (scanf("%ld",&n)>-1)
  {
    if (n==0) {printf("0\n");continue;}
    a.v[1][1]=1;a.v[2][1]=1;
    b.v[1][1]=0;b.v[1][2]=1;b.v[2][1]=1;b.v[2][2]=1;
    quick();
    printf("%ld",a.v[1][1]);
  }
  return 0;
}

#include
using namespace std;
const int mo=10000;
struct arr{int v[3][3];}a,b;
int n;
arr chen(arr x,arr y)
{
  arr s;int m=2,n=2,p=2;
  for (int i=1;i<=m;i++)
    for (int j=1;j<=p;j++)
      {
        s.v[i][j]=0;
        for (int k=1;k<=n;k++)
          s.v[i][j]=(s.v[i][j]+x.v[i][k]*y.v[k][j]%mo)%mo;
      }
  return s;
}
void quick()
{
  int mi=n-2;a=b;
  while (mi>0)
  {
    if (mi&1) a=chen(a,b);
    b=chen(b,b);
    mi/=2;
  }
}
int main()
{
  while (scanf("%ld",&n)>-1)
  {
    if (n==0) {printf("0\n");continue;}
    b.v[1][1]=1;b.v[1][2]=1;b.v[2][1]=1;b.v[2][2]=0;
    quick();
    printf("%ld\n",a.v[1][1]);
  }
  return 0;
}