斐波那契数（POJ 3070）

问题描述：斐波那契数列是指如下数列: F(0)=0, F(1)=1, F(n)=F(n-1)+F(n-2).数列的前10个数如下:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
对于给定的数n,请求出F(n)的后四个数字.

关于输入：输入包括多组数据.
每组数据为一行,包含一个非负整数n(0<=n<=1000000000).
输入的最后一行是一个数-1,表示输入结束.

 

关于输出：对于每组数据,你的程序应该输出F(n)对10000取模的结果.

 

例子输入：

0
9
999999999
1000000000
-1例子输出：

0
34
626
6875解题思路：利用分治法计算矩阵乘积
 

代码：

#include <iostream>#include <vector>using namespace std;//定义矩阵struct BMATRIX{long long f11;long long f12;long long f21;long long f22;};//矩阵乘法BMATRIX matMulti(BMATRIX mat1, BMATRIX mat2){BMATRIX matrix;matrix.f11 = (mat1.f11*mat2.f11 + mat1.f12*mat2.f21)%10000;matrix.f12 = (mat1.f11*mat2.f12 + mat1.f12*mat2.f22)%10000;matrix.f21 = (mat1.f21*mat2.f11 + mat1.f22*mat2.f21)%10000;matrix.f22 = (mat1.f21*mat2.f12 + mat1.f22*mat2.f22)%10000;return matrix; }BMATRIX Fibo(int m){BMATRIX unitMat,retMat;unitMat.f11 = (long long)1;unitMat.f12 = (long long)1;unitMat.f21 = (long long)1;unitMat.f22 = (long long)0;if(m==1){retMat.f11 = (long long)1;retMat.f12 = (long long)1;retMat.f21 = (long long)1;retMat.f22 = (long long)0;}else if(m%2 == 0){retMat = Fibo(m/2);retMat = matMulti(retMat,retMat);}else if(m%2 == 1){retMat = Fibo((m-1)/2);retMat = matMulti(retMat,retMat);retMat = matMulti(retMat,unitMat);}return retMat;}int main(){int m;vector<int> output;BMATRIX retMatrix;retMatrix.f11 = (long long)0;retMatrix.f12 = (long long)0;retMatrix.f21 = (long long)0;retMatrix.f22 = (long long)0;while( cin>>m && m!=-1){if(m==0){output.push_back(0);}else if(m==1){output.push_back(1);}else{retMatrix = Fibo(m);output.push_back( retMatrix.f12 % 10000 );}}for(vector<int>::size_type i=0; i<output.size(); i++)cout<<output[i]<<endl;system("pause");return 0;}