POJ_Fibonacci POJ_3070(矩阵快速幂入门题,附上自己写的矩阵模板)
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10521 | Accepted: 7477 |
Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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
.
Given an integer n, your goal is to compute the last 4 digits of Fn.
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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn 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
typedef struct Matrix
{
// Made by xiper , Last updata : 2015 / 6 / 14
int r , c , ele[][];
Matrix(const int & r , const int & c)
{
this->r = r , this->c = c; // i will not init for ele , u should do it
}
friend ostream& operator << (ostream & os,const Matrix & x)
{
for(int i = ; i < x.r ; ++ i)
{
for(int j = ; j < x.c ; ++ j)
os << x.ele[i][j] << " ";
os << endl;
}
return os;
}
Matrix operator * (const Matrix & x) const
{
if (c != x.r)
{
cout << "Error on Matrix operator * , (c1 != r1)" << endl;
return Matrix(,);
}
Matrix res(r,x.c);
for(int i = ; i < r ; ++ i)
for(int j = ; j < x.c ; ++ j)
{
int sum = ;
for(int k = ; k < c ; ++ k)
sum += ele[i][k]*x.ele[k][j];
res.ele[i][j] = sum;
}
return res;
}
Matrix operator * (const int & x ) const
{
Matrix res(r,c);
for(int i = ; i < r ; ++ i)
for(int j = ; j < c ; ++ j)
res.ele[i][j] = ele[i][j]*x;
return res;
}
Matrix operator + (const Matrix & x) const
{
if (x.r != r || x.c != c)
{
cout << "Error on Matrix operator + , (r1 != r2 || c1 != c2)" << endl;
return Matrix(,);
}
Matrix res(r,c);
for(int i = ; i < r ; ++ i)
for(int j = ; j < c ; ++ j)
res.ele[i][j] = ele[i][j] + x.ele[i][j];
return res;
}
Matrix operator - (const Matrix & x) const
{
if (x.r != r || x.c != c)
{
cout << "Error on Matrix operator + , (r1 != r2 || c1 != c2)" << endl;
return Matrix(,);
}
Matrix res(r,c);
for(int i = ; i < r ; ++ i)
for(int j = ; j < c ; ++ j)
res.ele[i][j] = ele[i][j] - x.ele[i][j];
return res;
}
void r_ope(int whichr , int num)
{
for(int i = ; i < c ; ++ i)
ele[whichr][i] += num;
}
void c_ope(int whichc , int num)
{
for(int i = ; i < r ; ++ i)
ele[i][whichc] += num;
}
void init(int x)
{
for(int i = ; i < r ; ++ i)
for(int j = ; j < c ; ++ j)
ele[i][j] = x;
}
void init_dig()
{
memset(ele,,sizeof(ele));
for(int i = ; i < min(r,c) ; ++ i)
ele[i][i] = ;
}
Matrix Mulite (const Matrix & x ,int mod) const
{
if (c != x.r)
{
cout << "Error on Matrix function Mulite(pow may be) , (c1 != r1)" << endl;
return Matrix(,);
}
Matrix res(r,x.c);
for(int i = ; i < r ; ++ i)
for(int j = ; j < x.c ; ++ j)
{
int sum = ;
for(int k = ; k < c ; ++ k)
sum += (ele[i][k]*x.ele[k][j]) % mod;
res.ele[i][j] = sum % mod;
}
return res;
}
Matrix pow(int n , int mod)
{
if (r != c)
{
cout << "Error on Matrix function pow , (r != c)" << endl;
return Matrix(,);
}
Matrix tmp(r,c);
memcpy(tmp.ele,ele,sizeof(ele));
Matrix res(r,c);
res.init_dig();
while(n)
{
if (n & )
res = res.Mulite(tmp,mod);
n >>= ;
tmp = tmp.Mulite(tmp,mod);
}
return res;
}
};
AC代码就不贴了(其实就几行。。)
POJ_Fibonacci POJ_3070(矩阵快速幂入门题,附上自己写的矩阵模板)的更多相关文章
- hdu 1575 Tr A (矩阵快速幂入门题)
题目 先上一个链接:十个利用矩阵乘法解决的经典题目 这个题目和第二个类似 由于矩阵乘法具有结合律,因此A^4 = A * A * A * A = (A*A) * (A*A) = A^2 * A^2.我 ...
- 矩阵快速幂(入门) 学习笔记hdu1005, hdu1575, hdu1757
矩阵快速幂是基于普通的快速幂的一种扩展,如果不知道的快速幂的请参见http://www.cnblogs.com/Howe-Young/p/4097277.html.二进制这个东西太神奇了,好多优秀的算 ...
- HDU 1575 Tr A 【矩阵经典2 矩阵快速幂入门】
任意门:http://acm.hdu.edu.cn/showproblem.php?pid=1575 Tr A Time Limit: 1000/1000 MS (Java/Others) Me ...
- HDU 1575 矩阵快速幂裸题
题意:中文题 我就不说了吧,... 思路:矩阵快速幂 // by SiriusRen #include <cstdio> #include <cstring> using na ...
- POJ3070矩阵快速幂简单题
题意: 求斐波那契后四位,n <= 1,000,000,000. 思路: 简单矩阵快速幂,好久没刷矩阵题了,先找个最简单的练练手,总结下矩阵推理过程,其实比较简单,关键 ...
- 集训第六周 矩阵快速幂 K题
Description In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. F ...
- Foj1683矩阵快速幂水题
Foj 1683 纪念SlingShot 题目链接:http://acm.fzu.edu.cn/problem.php?pid=1683 题目:已知 F(n)=3 * F(n-1)+2 * F(n-2 ...
- LightOJ 1065 - Number Sequence 矩阵快速幂水题
http://www.lightoj.com/volume_showproblem.php?problem=1065 题意:给出递推式f(0) = a, f(1) = b, f(n) = f(n - ...
- hdu 2604 Queuing(矩阵快速幂乘法)
Problem Description Queues and Priority Queues are data structures which are known to most computer ...
随机推荐
- poj3696:同余方程,欧拉定理
感觉很不错的数学题,可惜又是看了题解才做出来的 题目大意:给定一个数n,找到8888....(x个8)这样的数中,满足能整除n的最小的x,若永远无法整除n 则输出0 做了这个题和后面的poj3358给 ...
- HDU4821---字符串hash,map判重
这是2013年长春区域赛的铜牌题...然而第一次做的时候一直觉得会超时的..最后才知道并没有想象中的那么恐怖: 这题有两个注意的地方: (1)h[i] = h[i-1] * seed + s[i] - ...
- CSS的基本认识
1.定义: 级联样式表(Cascading Style Sheet)简称“CSS”,通常又称为“风格样式表(Style Sheet)”,它是用来进行网页风格设计的. 2.对CSS的基本认识: CSS是 ...
- install-file -Dfile=J:\project01\workspace\service\lib\javapns-jdk16-163.jar -DgroupId=org.json -Dar
今天在开发项目的时候发现了一个问题,所以通过博客来记录起来! 为了以后在问题的解决方面能得到借鉴! 问题的现象是这种: 这样会报错的.pom.xml文件他在编译.检查他的文件语法的时候是须要參考库中的 ...
- java基础之String
字符串的含义 字符串的应用 字符串的方法
- C# 模拟键盘按键操作
[DllImport("user32.dll")] public static extern IntPtr keybd_event(byte bVk, byte bScan, in ...
- c# 根据窗口截图,合并图片
c# 根据窗口截图,合并图片 public class CaptureWindows { #region 类 /// <summary> /// Helper class containi ...
- C#语法糖: 扩展方法(常用)
今天继续分享C#4.0语法糖的扩展方法,这个方法也是我本人比较喜欢的方法.大家先想想比如我们以前写的原始类型不能满足现在的需求,而需要在该类型中添加新的方法来实现时大家会怎么做.我先说一下我没有学习到 ...
- LNMP优化
LNMP优化 LNMP优化从系统安全,系统资源占用率,及web服务并发负载这三个方面体现,并 且主要体现在web服务并发负载这一方面. 1:首先进行linux优化加固 Linux ...
- silverlight+wcf 项目 silverlight获得web程序的参数
silverlight 可以通过属性InitParams 获得参数,如果参数是动态的需要web程序传递的,具体操作如下: web程序后台:AppID,USERID需要的参数 this.frmRepor ...