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:
.
纯模板题、- -
#include <iostream> #include <cstdio> #include <time.h> using namespace std; #define MOD 10000 #define N 2 void mul(int a[N][N],int b[N][N]) { int i,j,k; int c[N][N]={0}; for(i=0;i<N;i++) { for(j=0;j<N;j++) { for(k=0;k<N;k++) { c[i][j]=(c[i][j]+a[i][k]*b[k][j])%MOD; } } } for(i=0;i<N;i++) { for(j=0;j<N;j++) { a[i][j]=c[i][j]; } } } int main() { int n; while(scanf("%d",&n),n+1) { int a[N][N]={{0},{1}},b[N][N]={{1,1},{1,0}}; //首项和递推矩阵 while(n) //二分快速幂 { if(n&1) mul(a,b); mul(b,b); n>>=1; } printf("%d\n",a[1][1]); } return 0; }
矩阵快速幂 [POJ 3070 NYOJ 148] Fibonacci
原文：http://www.cnblogs.com/hate13/p/4123338.html