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| 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[50][50]; 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 = 0 ; i < x.r ; ++ i) { for(int j = 0 ; 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(0,0); } Matrix res(r,x.c); for(int i = 0 ; i < r ; ++ i) for(int j = 0 ; j < x.c ; ++ j) { int sum = 0; for(int k = 0 ; 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 = 0 ; i < r ; ++ i) for(int j = 0 ; 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(0,0); } Matrix res(r,c); for(int i = 0 ; i < r ; ++ i) for(int j = 0 ; 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(0,0); } Matrix res(r,c); for(int i = 0 ; i < r ; ++ i) for(int j = 0 ; 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 = 0 ; i < c ; ++ i) ele[whichr][i] += num; } void c_ope(int whichc , int num) { for(int i = 0 ; i < r ; ++ i) ele[i][whichc] += num; } void init(int x) { for(int i = 0 ; i < r ; ++ i) for(int j = 0 ; j < c ; ++ j) ele[i][j] = x; } void init_dig() { memset(ele,0,sizeof(ele)); for(int i = 0 ; i < min(r,c) ; ++ i) ele[i][i] = 1; } 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(0,0); } Matrix res(r,x.c); for(int i = 0 ; i < r ; ++ i) for(int j = 0 ; j < x.c ; ++ j) { int sum = 0; for(int k = 0 ; 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(0,0); } Matrix tmp(r,c); memcpy(tmp.ele,ele,sizeof(ele)); Matrix res(r,c); res.init_dig(); while(n) { if (n & 1) res = res.Mulite(tmp,mod); n >>= 1; 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)
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原文地址：http://www.cnblogs.com/Xiper/p/4575415.html
